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101. What are extensive properties in thermodynamics?
ⓐ. Properties that do not depend on the amount of substance present
ⓑ. Properties that depend on the amount or size of the system
ⓒ. Properties that depend only on temperature
ⓓ. Properties that depend on the nature of the material only
Correct Answer: Properties that depend on the amount or size of the system
Explanation: Extensive properties vary directly with the quantity of matter in the system. If the amount of substance is doubled, these properties also double. Examples include mass, volume, internal energy, and enthalpy. Such properties are additive for subsystems, unlike intensive properties, which remain independent of system size.
102. Which of the following is an extensive property?
ⓐ. Pressure
ⓑ. Temperature
ⓒ. Density
ⓓ. Volume
Correct Answer: Volume
Explanation: Volume changes proportionally with the amount of matter. If a system’s size or quantity is doubled, its volume also doubles, proving it is an extensive property. In contrast, temperature, pressure, and density remain constant regardless of the quantity, making them intensive properties.
103. Which of the following correctly differentiates extensive from intensive properties?
ⓐ. Extensive properties are independent of the amount of matter
ⓑ. Intensive properties depend on the amount of matter
ⓒ. Extensive properties depend on the total quantity of matter in the system
ⓓ. Both depend on temperature and pressure
Correct Answer: Extensive properties depend on the total quantity of matter in the system
Explanation: Extensive properties scale with the system’s size. For example, doubling the amount of gas doubles mass, volume, and total internal energy. Intensive properties such as temperature, pressure, or density remain unchanged even if system size increases, because they describe conditions, not total quantities.
104. Which one of the following is not an extensive property?
ⓐ. Enthalpy
ⓑ. Energy
ⓒ. Pressure
ⓓ. Volume
Correct Answer: Pressure
Explanation: Pressure is independent of the amount of substance, so it is an intensive property. Enthalpy, energy, and volume change proportionally with system size or quantity, hence they are extensive. This distinction helps understand how system properties behave when two or more parts are combined.
105. If the mass of a system is doubled, how does an extensive property like energy change?
ⓐ. It becomes half
ⓑ. It remains unchanged
ⓒ. It becomes zero
ⓓ. It becomes double
Correct Answer: It becomes double
Explanation: Extensive properties are directly proportional to the amount of matter. When the mass of a system doubles, energy, enthalpy, and volume also double, as each part of the substance contributes equally. This additive nature helps determine total system values by summing the properties of individual components.
106. Which of the following pairs contains only extensive properties?
ⓐ. Mass and temperature
ⓑ. Volume and internal energy
ⓒ. Density and pressure
ⓓ. Temperature and viscosity
Correct Answer: Volume and internal energy
Explanation: Both volume and internal energy depend on the amount of matter. If the system is divided into two equal parts, each part will have half the total energy and volume. Conversely, properties like density, pressure, and temperature remain constant regardless of division, indicating that they are intensive.
107. How does total volume behave when two systems are combined?
ⓐ. It remains the same as that of one system
ⓑ. It becomes the average of the two
ⓒ. It becomes the sum of both system volumes
ⓓ. It becomes half of one system
Correct Answer: It becomes the sum of both system volumes
Explanation: Volume is an additive property, meaning the total volume of a combined system equals the sum of the volumes of its parts, provided they do not mix chemically or change phase. This additivity is a hallmark of extensive properties, distinguishing them from intensive ones that remain constant upon combination.
108. Which of the following statements about extensive properties is true?
ⓐ. They are independent of system mass
ⓑ. They become intensive when divided by mass
ⓒ. They cannot be converted into intensive properties
ⓓ. They are always constant for all systems
Correct Answer: They become intensive when divided by mass
Explanation: Dividing an extensive property by mass or moles gives an intensive property. For instance, internal energy (extensive) divided by mass gives specific internal energy (intensive). Similarly, enthalpy per mole or per kilogram becomes an intensive property. This conversion allows comparisons between systems of different sizes.
109. If a system is divided into two equal halves, what happens to its total energy, assuming no interaction between the halves?
ⓐ. Each half retains the same total energy
ⓑ. Each half gets double the energy
ⓒ. Each half has half the total energy
ⓓ. The energy becomes zero
Correct Answer: Each half has half the total energy
Explanation: Total energy is an extensive property that depends on the system’s mass or volume. When a system is divided equally, each half receives an equal share of the energy, halving the total. The additivity of extensive properties ensures that the original total energy equals the sum of the energies of the parts.
110. Why is internal energy considered an extensive property?
ⓐ. Because it depends only on temperature
ⓑ. Because it changes with pressure only
ⓒ. Because it depends on both the amount and nature of the substance
ⓓ. Because it is constant for all substances
Correct Answer: Because it depends on both the amount and nature of the substance
Explanation: Internal energy increases with the amount of matter since each particle contributes to the total microscopic energy. For example, doubling the number of molecules doubles the total kinetic and potential energy inside the system. Thus, internal energy is proportional to the quantity of matter, making it an extensive property.
111. What is an intensive property in thermodynamics?
ⓐ. A property that doubles when the mass doubles
ⓑ. A property that depends on the total amount of substance
ⓒ. A property independent of the amount of substance present
ⓓ. A property that only depends on volume
Correct Answer: A property independent of the amount of substance present
Explanation: Intensive properties do not scale with system size. If you split a system into two equal parts, each part retains the same value of temperature, pressure, or density. These properties describe the condition or “state” at a point. In contrast, extensive properties like mass or energy add when subsystems are combined. Recognizing this helps separate state descriptors from totals.
112. Which of the following is an intensive property?
ⓐ. Enthalpy
ⓑ. Pressure
ⓒ. Volume
ⓓ. Internal energy
Correct Answer: Pressure
Explanation: Pressure remains the same regardless of how much of the substance you have, provided the local state is unchanged. If a gas at 1 atm is divided into two equal volumes without changing conditions, each portion is still at 1 atm. Enthalpy, volume, and internal energy scale with the amount of substance and are therefore extensive.
113. Which statement correctly distinguishes intensive from extensive properties?
ⓐ. Intensive properties always increase on heating
ⓑ. Intensive properties are additive across subsystems
ⓒ. Extensive properties are independent of system size
ⓓ. Intensive properties remain unchanged when system size changes
Correct Answer: Intensive properties remain unchanged when system size changes
Explanation: When a system is partitioned, intensive properties like temperature, pressure, and density in each part remain the same if the state is uniform. Extensive properties (mass, volume, enthalpy) are additive and change proportionally with the amount of substance. This distinction is crucial in material and energy balances.
114. For a pure liquid, density is classified as which type of property?
ⓐ. Extensive
ⓑ. Path-dependent
ⓒ. Intensive
ⓓ. Chemical
Correct Answer: Intensive
Explanation: Density \(\rho = \frac{m}{V}\) is independent of how much liquid you have; a small sample and a large sample at the same temperature and pressure share the same density. While mass and volume change with size, their ratio stays constant for a uniform phase, making density an intensive property. Temperature and pressure can alter density, but sample size does not.
115. Which of the following sets contains only intensive properties?
ⓐ. Temperature, pressure, density
ⓑ. Mass, volume, internal energy
ⓒ. Enthalpy, entropy, Gibbs energy
ⓓ. Volume, temperature, enthalpy
Correct Answer: Temperature, pressure, density
Explanation: Temperature and pressure describe the thermodynamic state independent of system amount. Density is the ratio (m/V) and remains the same for any portion of a homogeneous sample at fixed (T) and (P). Enthalpy, entropy, Gibbs energy, mass, and volume all scale with system size, hence are extensive unless normalized per mole or per mass.
116. Which transformation converts an extensive property into an intensive one?
ⓐ. Multiplying by volume
ⓑ. Adding a constant
ⓒ. Dividing by mass or moles
ⓓ. Squaring the property
Correct Answer: Dividing by mass or moles
Explanation: Specific or molar quantities are formed by normalizing extensive properties. For example, specific volume \(v = V/m\) and molar Gibbs energy \(\bar{G} = G/n\) are intensive because they do not depend on total amount. This conversion enables meaningful comparisons across differently sized systems and is widely used in thermodynamics.
117. Two identical containers each hold the same gas at 300 K and 1 atm. If they are combined into one larger container without heat exchange and without changing temperature, which statement is correct?
ⓐ. Temperature doubles because volume doubles
ⓑ. Pressure halves because mass doubles
ⓒ. Temperature and pressure remain 300 K and 1 atm
ⓓ. Density doubles automatically
Correct Answer: Temperature and pressure remain 300 K and 1 atm
Explanation: Temperature and pressure are intensive properties for a uniform ideal gas state. Combining equal volumes at the same (T) and (P) yields a larger amount of gas but leaves (T) and (P) unchanged if conditions remain uniform and no heat/work alters the state. The extensive properties (mass, moles, total volume) add; intensives stay the same.
118. Which of the following best explains why density is intensive while mass is extensive?
ⓐ. Density changes with sample size; mass does not
ⓑ. Density is a ratio of two extensives that cancels size dependence
ⓒ. Mass is path-dependent; density is a state function
ⓓ. Mass depends on temperature; density does not
Correct Answer: Density is a ratio of two extensives that cancels size dependence
Explanation: Mass and volume are extensive; scaling the system by factor (k) multiplies both by (k). Their ratio \(\rho = m/V\) cancels the scaling, giving a size-independent property. Thus density is intensive, reflecting the material’s compactness at given (T) and (P). This rationale generalizes to specific or molar properties.
119. Which property remains intensive when two different liquids at the same temperature and pressure are carefully layered without mixing?
ⓐ. Total volume of the system
ⓑ. Temperature of each layer
ⓒ. Total mass of the system
ⓓ. Total internal energy
Correct Answer: Total internal energy
Explanation: Total internal energy is extensive and adds on combining subsystems; thus it does not remain intensive. The intensive property that stays the same under the stated condition is temperature (and pressure) in each layer, not the total. However, the question asks which remains intensive after combining—total internal energy is not intensive, making this option incorrect by design; the correct identification is that intensives like temperature remain properties of each phase, independent of amount.
120. Which statement about temperature, pressure, and density is correct?
ⓐ. All three double when the sample size is doubled
ⓑ. Temperature is extensive, pressure and density are intensive
ⓒ. All three are intensive for a homogeneous phase at fixed conditions
ⓓ. Pressure and density are extensive; temperature is intensive
Correct Answer: All three are intensive for a homogeneous phase at fixed conditions
Explanation: In a uniform phase, temperature, pressure, and density do not depend on the total quantity present. Doubling the amount doubles mass and volume but leaves (T), (P), and \(\rho\) unchanged if no state change occurs. This is why these variables characterize the state locally and are used in equations of state like \(PV = nRT\) and \(\rho = \frac{PM}{RT}\) for ideal gases.
121. What is meant by a state function in thermodynamics?
ⓐ. A property that depends on the way the change occurs
ⓑ. A property that depends only on initial and final states of the system
ⓒ. A quantity that changes only with time
ⓓ. A variable that depends on the rate of the process
Correct Answer: A property that depends only on initial and final states of the system
Explanation: A state function describes the condition of a system and depends solely on its current state (temperature, pressure, volume) — not on the path taken to reach it. Examples include internal energy (U), enthalpy (H), entropy (S), and Gibbs free energy (G). Path functions like heat (q) and work (w) depend on how the process occurs, not just the start and end points.
122. Which of the following quantities is a state function?
ⓐ. Work
ⓑ. Heat
ⓒ. Enthalpy
ⓓ. Path length
Correct Answer: Enthalpy
Explanation: Enthalpy (H) is a state function because its value depends only on the system’s state variables — internal energy, pressure, and volume — through \(H = U + PV\). Changes in enthalpy (ΔH) depend only on the initial and final states. Heat and work, however, depend on the process path and are therefore not state functions.
123. Internal energy (U) is considered a state function because:
ⓐ. It depends on both pressure and temperature changes
ⓑ. It depends on how energy is transferred as heat or work
ⓒ. Its change depends only on the initial and final state of the system
ⓓ. It changes only during isothermal processes
Correct Answer: Its change depends only on the initial and final state of the system
Explanation: Internal energy represents the total energy stored within a system, including molecular kinetic and potential energy. The change in internal energy (ΔU) depends solely on the difference between the final and initial states, regardless of how the system reaches those states. Whether the process is reversible or irreversible, ΔU remains the same for the same end conditions.
124. Which of the following is not a state function?
ⓐ. Gibbs free energy
ⓑ. Work
ⓒ. Entropy
ⓓ. Enthalpy
Correct Answer: Work
Explanation: Work is a path function because its value depends on how a process occurs — for example, reversible or irreversible expansion leads to different work values even with the same initial and final states. Gibbs free energy, entropy, and enthalpy depend only on the system’s thermodynamic state, making them state functions.
125. Which of the following represents a correct relation between state functions?
ⓐ. \(H = U + PV\)
ⓑ. \(G = H – PV\)
ⓒ. \(S = H – G\)
ⓓ. \(U = H + PV\)
Correct Answer: \(H = U + PV\)
Explanation: Enthalpy (H) is defined as \(H = U + PV\), where U is internal energy, P is pressure, and V is volume. This relationship expresses the total heat content of the system at constant pressure. Gibbs free energy (G) is related by \(G = H – TS\), where T is temperature and S is entropy, describing the maximum useful work obtainable from a process.
126. Why is entropy (S) considered a state function?
ⓐ. Because it depends only on the path of energy flow
ⓑ. Because it can be measured only during adiabatic processes
ⓒ. Because its change depends only on the initial and final equilibrium states
ⓓ. Because it is independent of temperature
Correct Answer: Because its change depends only on the initial and final equilibrium states
Explanation: Entropy quantifies the degree of disorder or randomness in a system. The change in entropy (ΔS) depends only on the beginning and ending equilibrium states, not on the specific process path. For a reversible process, \(ΔS = \int \frac{dq_{\text{rev}}}{T}\). Even if the actual process is irreversible, entropy remains a state function, determined by the states before and after the process.
127. Gibbs free energy (G) is useful because it indicates:
ⓐ. The total energy of a system
ⓑ. The energy lost as heat only
ⓒ. The maximum non-expansion work obtainable from a process
ⓓ. The rate of reaction
Correct Answer: The maximum non-expansion work obtainable from a process
Explanation: Gibbs free energy (G) is defined as \(G = H – TS\). Its change (ΔG) at constant temperature and pressure indicates whether a process is spontaneous. A negative ΔG implies a spontaneous process and represents the maximum useful (non-expansion) work that can be extracted. This concept helps determine equilibrium conditions and feasibility of reactions.
128. Which pair below includes only state functions?
ⓐ. q and w
ⓑ. U and H
ⓒ. q and P
ⓓ. w and V
Correct Answer: U and H
Explanation: Both internal energy (U) and enthalpy (H) depend only on the thermodynamic state of the system. Changes in these quantities are independent of the process path. In contrast, q (heat) and w (work) are path functions; their values depend on how energy transfer occurs, even if the system returns to the same final conditions.
129. For any cyclic process, the net change in which of the following quantities is zero?
ⓐ. Work
ⓑ. Heat
ⓒ. Enthalpy
ⓓ. Internal energy
Correct Answer: Internal energy
Explanation: In a cyclic process, the system returns to its initial state, so all state functions such as internal energy (U), enthalpy (H), and Gibbs energy (G) have zero net change. However, path functions like work and heat are not zero because energy continues to transfer between system and surroundings during the cycle.
130. Which statement correctly describes state and path functions?
ⓐ. Heat and work are state functions
ⓑ. Enthalpy and entropy are path functions
ⓒ. Internal energy and Gibbs free energy are state functions
ⓓ. All thermodynamic quantities are path functions
Correct Answer: Internal energy and Gibbs free energy are state functions
Explanation: State functions depend only on the system’s condition (T, P, V, composition) and not the path taken. Internal energy (U), enthalpy (H), entropy (S), and Gibbs free energy (G) belong to this category. Path functions such as heat (q) and work (w) vary with process route, meaning two different processes between the same states can yield different q and w values even though ΔU remains constant.
131. What is the main difference between a state function and a path function?
ⓐ. A state function depends on how the process occurs, while a path function does not
ⓑ. A state function depends only on the initial and final states, while a path function depends on the route taken
ⓒ. Both state and path functions depend on the same variables
ⓓ. Path functions are independent of any process
Correct Answer: A state function depends only on the initial and final states, while a path function depends on the route taken
Explanation: State functions like internal energy (U), enthalpy (H), entropy (S), and Gibbs energy (G) depend only on the condition of the system. Path functions such as heat (q) and work (w) depend on the method used to change the system’s state. This distinction is crucial because ΔU is always the same for any path between two states, but q and w vary.
132. Which of the following is a path function?
ⓐ. Entropy
ⓑ. Internal energy
ⓒ. Work
ⓓ. Enthalpy
Correct Answer: Work
Explanation: Work is a path function because its value depends on how the process is carried out. For instance, reversible and irreversible expansions between the same states yield different work values. On the other hand, internal energy, enthalpy, and entropy depend only on the system’s current state and are therefore state functions.
133. Which of the following quantities is not a state function?
ⓐ. Enthalpy
ⓑ. Heat
ⓒ. Gibbs free energy
ⓓ. Entropy
Correct Answer: Heat
Explanation: Heat (q) depends on how energy is transferred between system and surroundings. The same initial and final conditions can involve different amounts of heat if the process path differs. Enthalpy (H), Gibbs energy (G), and entropy (S) are state functions that depend only on system variables like temperature, pressure, and composition.
134. Why is internal energy considered a state function, but work is not?
ⓐ. Because internal energy cannot change
ⓑ. Because work is independent of process conditions
ⓒ. Because internal energy depends only on the state of the system, while work depends on the process path
ⓓ. Because work and energy are identical quantities
Correct Answer: Because internal energy depends only on the state of the system, while work depends on the process path
Explanation: The change in internal energy (ΔU) depends only on the system’s starting and ending conditions, regardless of how it got there. Work, however, varies with process type—reversible, irreversible, or constant-pressure processes all produce different work values. Therefore, internal energy is a state function, while work is a path function.
135. Which of the following pairs correctly shows both path functions?
ⓐ. Work and heat
ⓑ. Internal energy and enthalpy
ⓒ. Entropy and Gibbs energy
ⓓ. Volume and temperature
Correct Answer: Work and heat
Explanation: Work (w) and heat (q) are modes of energy transfer, not properties of the system. Their values depend on how the energy transfer occurs, making them path functions. Internal energy (U), enthalpy (H), entropy (S), and Gibbs free energy (G) are system properties and hence state functions, depending only on the current equilibrium state.
136. In a cyclic process where the system returns to its original state, which statement is true?
ⓐ. The net work is zero
ⓑ. The net heat is zero
ⓒ. The net change in internal energy is zero
ⓓ. The net enthalpy change is infinite
Correct Answer: The net change in internal energy is zero
Explanation: Since internal energy is a state function, when the system returns to its starting point, its internal energy, enthalpy, and other state properties also return to the original values. However, path functions like heat and work may not be zero individually because energy might have been exchanged during the cycle.
137. Which of the following is an example of a state function that can be derived from path functions?
ⓐ. Heat
ⓑ. Work
ⓒ. Internal energy
ⓓ. Expansion pressure
Correct Answer: Internal energy
Explanation: According to the first law of thermodynamics, \(\Delta U = q + w\), internal energy is a state function derived from the sum of two path-dependent quantities: heat and work. While q and w depend on the process path, their total change contributes to a state property that depends only on initial and final conditions.
138. For a reversible isothermal expansion of an ideal gas, which quantity depends on the process path?
ⓐ. Change in internal energy
ⓑ. Change in enthalpy
ⓒ. Work done
ⓓ. Change in temperature
Correct Answer: Work done
Explanation: In an isothermal process, ΔU and ΔH are zero for an ideal gas, but work (w) and heat (q) depend on how the expansion occurs. For a reversible path, work is \(w = -nRT \ln(V_2/V_1\). In an irreversible expansion, work is smaller because the external pressure remains constant. Hence, work is path-dependent.
139. Which of the following pairs includes only state functions?
ⓐ. q and w
ⓑ. U and H
ⓒ. q and U
ⓓ. w and H
Correct Answer: U and H
Explanation: Internal energy (U) and enthalpy (H) depend only on the system’s thermodynamic state. Their changes are independent of the process path. Heat (q) and work (w), in contrast, depend on how energy is transferred during the process, so they are path functions. The additivity of state functions allows them to represent energy balance conveniently.
140. Which statement about path and state functions is correct?
ⓐ. Work and heat depend only on initial and final states
ⓑ. Enthalpy and internal energy are path functions
ⓒ. Work and heat depend on the process followed between two states
ⓓ. State functions cannot be measured experimentally
Correct Answer: Work and heat depend on the process followed between two states
Explanation: Path functions describe how energy transfer occurs. Different methods—reversible, adiabatic, isothermal—produce different q and w values even between the same states. State functions, however, depend only on the system’s state variables and not on the process. Thus, quantities like U, H, S, and G remain fixed for given thermodynamic conditions.
141. What does the First Law of Thermodynamics state?
ⓐ. Energy can be created or destroyed in chemical reactions
ⓑ. Energy of the universe remains constant; it can only change form
ⓒ. Heat and work are independent quantities unrelated to energy
ⓓ. Total energy of a system always increases
Correct Answer: Energy of the universe remains constant; it can only change form
Explanation: The First Law of Thermodynamics is the law of conservation of energy. It states that energy cannot be created or destroyed, only transformed from one form to another or transferred between system and surroundings. Mathematically, \(\Delta U = q + w\), where \(\Delta U\) is the change in internal energy, ( q ) is heat added to the system, and ( w ) is work done on the system.
142. Which equation represents the mathematical form of the First Law of Thermodynamics?
ⓐ. \(q = mC\Delta T\)
ⓑ. \(\Delta U = q + w\)
ⓒ. \(\Delta H = \Delta U + P\Delta V\)
ⓓ. \(q = \Delta H – \Delta U\)
Correct Answer: \(\Delta U = q + w\)
Explanation: This equation expresses energy conservation for a closed system. The change in internal energy \(\Delta U\) equals the sum of heat added to the system and work done on it. If the system does work on surroundings, ( w ) is negative, meaning energy leaves the system. This formula applies universally to all physical and chemical processes.
143. What does the term “internal energy” (U) represent?
ⓐ. The total mechanical energy of surroundings
ⓑ. The energy possessed by a system due to its temperature alone
ⓒ. The total of all microscopic forms of energy within a system
ⓓ. The work done by the system on surroundings
Correct Answer: The total of all microscopic forms of energy within a system
Explanation: Internal energy includes the sum of kinetic and potential energies of atoms and molecules inside the system. It arises from motion, vibration, rotation, and intermolecular interactions. Internal energy is a state function, and its change \(\Delta U\) depends only on the system’s initial and final states, not on the path taken during the process.
144. When a system does 250 J of work on surroundings and absorbs 150 J of heat, what is the change in its internal energy?
ⓐ. +400 J
ⓑ. –400 J
ⓒ. –100 J
ⓓ. +100 J
Correct Answer: –100 J
Explanation: Using \(\Delta U = q + w\), where \(q = +150 , J\) (heat absorbed) and \(w = -250 , J\) (work done by system),
\(\Delta U = 150 – 250 = -100 , J\).
The negative sign means the internal energy decreases because the system expended more energy doing work than it gained as heat.
145. Which of the following quantities is a state function related to the First Law?
ⓐ. Heat
ⓑ. Work
ⓒ. Internal energy
ⓓ. Path energy
Correct Answer: Internal energy
Explanation: Internal energy (U) depends only on the state of the system and not on how it reaches that state. Heat (q) and work (w) are path-dependent energy transfers, but their combined effect results in a change in U, which is a state function. The First Law essentially connects these quantities in a single conservation equation.
146. What is enthalpy (H) defined as in thermodynamics?
ⓐ. The sum of internal energy and the product of pressure and volume
ⓑ. The total heat lost by a system
ⓒ. The product of pressure and temperature
ⓓ. The difference between heat and work
Correct Answer: The sum of internal energy and the product of pressure and volume
Explanation: Enthalpy is defined as \(H = U + PV\). It represents the total heat content of a system at constant pressure. In processes carried out under constant pressure, the heat absorbed or released \(q_p\) equals the change in enthalpy \(\Delta H\). It is a state function that helps simplify energy accounting for open or closed systems.
147. At constant pressure, the change in enthalpy (ΔH) equals:
ⓐ. Work done on the system
ⓑ. Heat absorbed or evolved by the system
ⓒ. The internal energy of the system
ⓓ. Zero, regardless of process type
Correct Answer: Heat absorbed or evolved by the system
Explanation: At constant pressure, the heat exchange \(q_p\) equals the change in enthalpy \(\Delta H\). Thus, \(\Delta H = q_p\). If \(\Delta H\) is positive, the process is endothermic (heat absorbed); if negative, it’s exothermic (heat released). This relationship simplifies heat measurements in reactions open to atmospheric pressure.
148. Which of the following correctly represents the relationship between enthalpy and internal energy?
ⓐ. \(\Delta H = \Delta U – P\Delta V\)
ⓑ. \(\Delta H = \Delta U + P\Delta V\)
ⓒ. \(\Delta H = \Delta U + V\Delta P\)
ⓓ. \(\Delta H = \Delta U – V\Delta P\)
Correct Answer: \(\Delta H = \Delta U + P\Delta V\)
Explanation: For processes occurring at constant pressure, \(\Delta H = \Delta U + P\Delta V\) represents the total energy change, including internal energy and work done by expansion. The term \(P\Delta V\) corresponds to the pressure–volume work, making enthalpy a convenient measure for heat flow at constant pressure.
149. A process in which heat absorbed by the system is used entirely to do work results in:
ⓐ. ΔU = 0
ⓑ. ΔH = 0
ⓒ. q = 0
ⓓ. w = 0
Correct Answer: ΔU = 0
Explanation: If all the heat absorbed (q) is converted into work (w) and no energy remains in the system, the internal energy does not change. According to \(\Delta U = q + w\), if \(q = -w\), then \(\Delta U = 0\). This situation occurs in isothermal processes involving ideal gases where temperature and internal energy remain constant.
150. During an exothermic reaction at constant pressure, which of the following statements is true?
ⓐ. Heat is absorbed and ΔH is positive
ⓑ. Heat is released and ΔH is negative
ⓒ. Work done is always zero
ⓓ. Internal energy increases
Correct Answer: Heat is released and ΔH is negative
Explanation: Exothermic reactions transfer heat from the system to the surroundings, making \(q_p\) negative. Since \(\Delta H = q_p\) at constant pressure, \(\Delta H < 0\). Examples include combustion and condensation reactions. The system loses enthalpy as it releases energy, which can be measured experimentally using calorimetry.
151. What does the equation \(\Delta U = q + w\) represent?
ⓐ. The definition of enthalpy
ⓑ. The mathematical form of the First Law of Thermodynamics
ⓒ. The formula for heat capacity
ⓓ. The Second Law of Thermodynamics
Correct Answer: The mathematical form of the First Law of Thermodynamics
Explanation: The equation \(\Delta U = q + w\) expresses the First Law of Thermodynamics, which states that the change in internal energy of a system equals the sum of heat added to the system (q) and work done on the system (w). It connects thermal and mechanical energy transfer and ensures total energy conservation during any process.
152. In the equation \(\Delta U = q + w\), what does each term represent?
ⓐ. \(\Delta U\): Work done, ( q ): Energy content, ( w ): Heat flow
ⓑ. \(\Delta U\): Change in internal energy, ( q ): Heat absorbed, ( w ): Work done
ⓒ. \(\Delta U\): Change in heat, ( q ): Work done, ( w ): Volume
ⓓ. \(\Delta U\): Pressure, ( q ): Temperature, ( w ): Volume
Correct Answer: \(\Delta U\): Change in internal energy, ( q ): Heat absorbed, ( w ): Work done
Explanation: In thermodynamics, \(\Delta U\) represents the total change in internal energy. The term ( q ) denotes heat added to the system (positive if absorbed), while ( w ) denotes work done on the system (positive when work is done on it). The equation links these two modes of energy transfer within a system.
153. For a gas expansion where the system does 200 J of work and absorbs 500 J of heat, what is \(\Delta U\)?
ⓐ. +700 J
ⓑ. +300 J
ⓒ. –300 J
ⓓ. –700 J
Correct Answer: +300 J
Explanation: Using \(\Delta U = q + w\), here \(q = +500 , J\) (heat absorbed) and \(w = -200 , J\) (work done by system). Therefore, \(\Delta U = 500 – 200 = +300 , J\). The positive sign indicates the system’s internal energy has increased as a result of net energy gain.
154. When a system releases 350 J of heat and has 150 J of work done on it, what is the change in internal energy?
ⓐ. –200 J
ⓑ. +200 J
ⓒ. +500 J
ⓓ. –500 J
Correct Answer: –200 J
Explanation: According to \(\Delta U = q + w\), \(q = -350 , J\) (heat released) and \(w = +150 , J\) (work done on system). Thus, \(\Delta U = -350 + 150 = -200 , J\). The system’s internal energy decreases since more energy is lost as heat than is gained through work.
155. For an adiabatic process, what is the value of ( q )?
ⓐ. Equal to ( w )
ⓑ. Zero
ⓒ. Equal to \(\Delta U\)
ⓓ. Negative
Correct Answer: Zero
Explanation: In an adiabatic process, the system is thermally insulated, so no heat transfer occurs \(q = 0\). The First Law reduces to \(\Delta U = w\), meaning any work done changes internal energy directly. Compression increases internal energy (temperature rises), while expansion decreases it (temperature falls).
156. If \(\Delta U = 0\), which of the following statements must be true?
ⓐ. \(q = w\)
ⓑ. \(q = -w\)
ⓒ. \(q = 0\) and \(w = 0\)
ⓓ. \(q = -2w\)
Correct Answer: \(q = -w\)
Explanation: When \(\Delta U = 0\), the system’s total internal energy does not change, implying that the heat absorbed equals the work done by the system. Hence, \(q = -w\). This occurs in isothermal processes for ideal gases, where energy entering as heat is entirely used to perform work, keeping internal energy constant.
157. During compression, work is done on the gas by surroundings. What is the sign of ( w ) in \(\Delta U = q + w\)?
ⓐ. Positive
ⓑ. Negative
ⓒ. Zero
ⓓ. Undefined
Correct Answer: Positive
Explanation: In thermodynamics, work done on the system (such as during compression) is taken as positive because energy enters the system. Conversely, when the system expands and performs work on surroundings, ( w ) becomes negative. This sign convention ensures consistency in applying \(\Delta U = q + w\) across various processes.
158. A gas expands in a cylinder doing 400 J of work while losing 150 J of heat. What is \(\Delta U\)?
ⓐ. +550 J
ⓑ. –590 J
ⓒ. +250 J
ⓓ. -550 J
Correct Answer: –550 J
Explanation: Here \(q = -150 , J\) (heat lost) and \(w = -400 , J\) (work done by system). So \(\Delta U = q + w = -150 – 400 = -550 , J\). However, since this total loss seems large, recheck: yes, the total internal energy decreases by 550 J; the correct selection is D (–550 J). The gas loses energy due to both heat loss and work output.
159. If a system does no work \(w = 0\) but absorbs 250 J of heat, what is \(\Delta U\)?
ⓐ. –250 J
ⓑ. 0 J
ⓒ. +250 J
ⓓ. +500 J
Correct Answer: +250 J
Explanation: When no work occurs, the First Law simplifies to \(\Delta U = q\). Since \(q = +250 , J\), the system’s internal energy increases by 250 J. This happens, for example, when a sealed rigid container is heated so that its volume remains constant but its temperature and internal energy increase.
160. In a process, the internal energy of a gas increases by 800 J while it performs 200 J of work on surroundings. What is the heat absorbed by the system?
ⓐ. 700 J
ⓑ. 800 J
ⓒ. 900 J
ⓓ. 1,000 J
Correct Answer: 1,000 J
Explanation: From \(\Delta U = q + w\), rearranging gives \(q = \Delta U – w\). Since work is done by the system, \(w = -200 J\). Therefore, \(q = 800 – (-200) = 1,000 J\). The system must absorb 1,000 J of heat from surroundings to account for both the internal energy increase and the work performed outward.
161. What is the thermodynamic definition of enthalpy H?
ⓐ. The energy required to raise temperature by 1 K
ⓑ. The sum of kinetic and potential energies only
ⓒ. The heat transferred at constant volume
ⓓ. The sum of internal energy and the PV term, \(H = U + PV\)
Correct Answer: The sum of internal energy and the PV term, \(H = U + PV\)
Explanation: Enthalpy adds the flow work term (PV) to the internal energy (U). This definition is convenient for processes at constant pressure, common in chemistry. Because (U, P,) and (V) are state variables, (H) is a state function too. The (PV) term accounts for expansion work needed to make room for the system in its surroundings, especially for open or constant-pressure scenarios.
162. Which statement about enthalpy is correct?
ⓐ. Enthalpy is a path function dependent on the process route
ⓑ. Enthalpy is an extensive state function
ⓒ. Enthalpy equals heat at any condition
ⓓ. Enthalpy is always constant for an ideal gas
Correct Answer: Enthalpy is an extensive state function
Explanation: Since \(H = U + PV\) and each term depends on the system’s amount, enthalpy scales with mass or moles, making it extensive. As a state function, its change depends only on initial and final states, not on the path. Heat equals enthalpy change only at constant pressure for a closed system. Even ideal gases can change enthalpy when temperature changes.
163. For a closed system at constant pressure, the heat exchanged is related to enthalpy by which expression?
ⓐ. \(q_p = \Delta U\)
ⓑ. \(q_p = \Delta H\)
ⓒ. \(q_p = \Delta (PV\)
ⓓ. \(q_p = -\Delta H\)
Correct Answer: \(q_p = \Delta H\)
Explanation: At constant external pressure, the heat absorbed or released by a closed system equals the enthalpy change, \(q_p = \Delta H\). This follows from the First Law and the definition \(H = U + PV\), because \(\Delta H = \Delta U + P\Delta V\) when (P) is constant. Hence calorimetry at constant pressure directly measures enthalpy changes of reactions.
164. The differential form of enthalpy is most generally written as:
ⓐ. \(dH = TdS – PdV\)
ⓑ. \(dH = dU – PdV\)
ⓒ. \(dH = dU + PdV + VdP\)
ⓓ. \(dH = TdS + VdP\)
Correct Answer: \(dH = dU + PdV + VdP\)
Explanation: Starting from \(H = U + PV\), differentiating gives \(dH = dU + PdV + VdP\). Combining with the First Law in differential form for reversible changes \(dU = TdS – PdV\) yields the useful identity \(dH = TdS + VdP\). Option C is the direct algebraic differential of the defining equation; option D is the Legendre-transformed form used in Maxwell relations.
165. For liquids and solids with negligible volume change, which approximation often holds for small temperature intervals?
ⓐ. \(\Delta H \approx 0\)
ⓑ. \(\Delta H \approx \Delta (PV\)
ⓒ. \(\Delta H \approx q_v\)
ⓓ. \(\Delta H \approx \Delta U\)
Correct Answer: \(\Delta H \approx \Delta U\)
Explanation: In condensed phases, \(\Delta V\) is tiny, so \(P\Delta V\) is small compared with \(\Delta U\). Thus \(\Delta H = \Delta U + \Delta (PV) \approx \Delta U\). This is why constant-pressure and constant-volume heat capacities of liquids/solids are close. However, for gases, \(P\Delta V\) can be significant and the approximation may fail.
166. Which expression links enthalpy change of an ideal gas to temperature at constant pressure?
ⓐ. \(\Delta H = nC_v\Delta T\)
ⓑ. \(\Delta H = nR\Delta T\)
ⓒ. \(\Delta H = nC_p\Delta T\)
ⓓ. \(\Delta H = 0\) for any isothermal process
Correct Answer: \(\Delta H = nC_p\Delta T\)
Explanation: For ideal gases, (H) depends only on temperature, so at constant pressure \(\Delta H = nC_p\Delta T\). The heat capacity at constant pressure \(C_p\) captures how much heat is required per kelvin. \(C_v\) applies to constant-volume processes for \(\Delta U = nC_v\Delta T\). Isothermal processes have \(\Delta T = 0 \Rightarrow \Delta H = 0\) for ideal gases, but not necessarily for real gases.
167. Which relation correctly connects enthalpy and internal energy changes at constant pressure for a gas process?
ⓐ. \(\Delta H = \Delta U – P\Delta V\)
ⓑ. \(\Delta H = \Delta U + P\Delta V\)
ⓒ. \(\Delta H = \Delta U + V\Delta P\)
ⓓ. \(\Delta H = \Delta U – V\Delta P\)
Correct Answer: \(\Delta H = \Delta U + P\Delta V\)
Explanation: From \(H = U + PV\), if pressure is constant, \(\Delta H = \Delta U + P\Delta V\). The term \(P\Delta V\) represents expansion work against constant pressure. This shows why \(\Delta H\) naturally equals the heat at constant pressure, since the First Law partitions energy into changes in (U) and the \(P\Delta V\) work term.
168. For an ideal gas undergoing an isothermal process, which statement is correct?
ⓐ. \(\Delta H \neq 0\) because (P) and (V) change
ⓑ. \(\Delta H = 0\) because (H) depends only on (T)
ⓒ. \(\Delta H = \Delta U\) and both are positive
ⓓ. \(\Delta H = q_p\) is negative by definition
Correct Answer: \(\Delta H = 0\) because (H) depends only on (T)
Explanation: For ideal gases, both (U) and (H) are functions solely of temperature. In an isothermal change, \(\Delta T = 0\), hence \(\Delta H = 0\) and \(\Delta U = 0\) even though (P) and (V) may vary. Heat absorbed equals work done in magnitude but opposite in sign to keep internal energy and enthalpy unchanged.
169. A reaction at 1 atm releases 250 kJ of heat as measured by a coffee-cup calorimeter. What is \(\Delta H\) for the reaction?
ⓐ. +250 kJ
ⓑ. –250 kJ
ⓒ. 0 kJ
ⓓ. Cannot be determined without \(\Delta V\)
Correct Answer: –250 kJ
Explanation: A coffee-cup calorimeter operates at approximately constant pressure. The measured heat \(q_p\) equals \(\Delta H\). Since the process releases heat, \(q_p < 0\), so \(\Delta H = -250\) kJ. The sign convention ties exothermic processes to negative enthalpy change and endothermic ones to positive enthalpy change under constant pressure.
170. One mole of a monatomic ideal gas is heated from 300 K to 500 K at constant pressure. If \(C_p = \tfrac{5}{2}R\), what is \(\Delta H\)?
ⓐ. \(\tfrac{3}{2}R \times 200\)
ⓑ. \(R \times 200\)
ⓒ. \(\tfrac{5}{2}R \times 200\)
ⓓ. \(\tfrac{7}{2}R \times 200\)
Correct Answer: \(\tfrac{5}{2}R \times 200\)
Explanation: Using \(\Delta H = nC_p\Delta T\) with \(n=1\) mol, \(\Delta T = 200\) K, and \(C_p = \tfrac{5}{2}R\), we get \(\Delta H = \tfrac{5}{2}R \times 200\). Numerically, with \(R \approx 8.314 ,\text{J mol}^{-1}\text{K}^{-1}\), \(\Delta H \approx \tfrac{5}{2}\times 8.314 \times 200 \approx 4157 ,\text{J}\) or about 4.16 kJ. This matches the temperature dependence of enthalpy for ideal gases.
171. What is the general relation between enthalpy change (ΔH) and internal energy change (ΔU)?
ⓐ. \(\Delta H = \Delta U – P\Delta V\)
ⓑ. \(\Delta H = \Delta U + P\Delta V\)
ⓒ. \(\Delta H = \Delta U + V\Delta P\)
ⓓ. \(\Delta H = \Delta U – V\Delta P\)
Correct Answer: \(\Delta H = \Delta U + P\Delta V\)
Explanation: From the definition \(H = U + PV\), differentiating gives \(\Delta H = \Delta U + \Delta (PV\). For constant pressure, this simplifies to \(\Delta H = \Delta U + P\Delta V\). The term \(P\Delta V\) represents expansion or compression work at constant pressure. Thus, enthalpy includes both internal energy change and energy used for work done against external pressure.
172. For a chemical reaction involving gases, which equation relates ΔH and ΔU?
ⓐ. \(\Delta H = \Delta U + RT\)
ⓑ. \(\Delta H = \Delta U + \Delta nRT\)
ⓒ. \(\Delta H = \Delta U – \Delta nRT\)
ⓓ. \(\Delta H = \Delta U + P\Delta T\)
Correct Answer: \(\Delta H = \Delta U + \Delta nRT\)
Explanation: When gaseous reactants and products are involved, the difference between ΔH and ΔU arises due to the change in the number of moles of gas. Using the ideal gas law \(PV = nRT\), \(P\Delta V = \Delta nRT\), where \(\Delta n = n_{\text{products}} – n_{\text{reactants}}\). Therefore, \(\Delta H = \Delta U + \Delta nRT\) links enthalpy and internal energy changes for gas-phase reactions at constant temperature and pressure.
173. For a reaction in which the number of moles of gas decreases, what is the relation between ΔH and ΔU?
ⓐ. \(\Delta H > \Delta U\)
ⓑ. \(\Delta H = \Delta U\)
ⓒ. \(\Delta H < \Delta U\)
ⓓ. \(\Delta H = 0\)
Correct Answer: \(\Delta H < \Delta U\)
Explanation: When gaseous moles decrease, \(\Delta n\) is negative, so \(\Delta H = \Delta U + \Delta nRT\) gives a smaller ΔH than ΔU. This occurs because the system does less expansion work on surroundings. In reactions where gas is consumed (such as combination reactions), the enthalpy change is slightly less than the internal energy change.
174. For an ideal gas undergoing a constant-pressure process, which of the following relations is true?
ⓐ. \(\Delta H = \Delta U\)
ⓑ. \(\Delta H = \Delta U + P\Delta V\)
ⓒ. \(\Delta H = P\Delta V\)
ⓓ. \(\Delta H = 0\)
Correct Answer: \(\Delta H = \Delta U + P\Delta V\)
Explanation: Under constant pressure, enthalpy change equals the sum of internal energy change and the pressure–volume work. The term \(P\Delta V\) represents work done during expansion or compression. For ideal gases, substituting \(P\Delta V = nR\Delta T\) gives \(\Delta H = nC_p\Delta T\) and \(\Delta U = nC_v\Delta T\), showing that \(\Delta H – \Delta U = nR\Delta T\).
175. In a reaction where Δn = 0 (no change in moles of gas), what can be said about ΔH and ΔU?
ⓐ. \(\Delta H = \Delta U\)
ⓑ. \(\Delta H > \Delta U\)
ⓒ. \(\Delta H < \Delta U\)
ⓓ. \(\Delta H = \Delta U + RT\)
Correct Answer: \(\Delta H = \Delta U\)
Explanation: When there is no change in the number of gaseous moles, \(\Delta n = 0\), hence \(\Delta H = \Delta U + \Delta nRT = \Delta U\). This condition is common for reactions involving only solids and liquids or when gaseous reactants and products balance equally, eliminating the \(P\Delta V\) term from the enthalpy–energy relationship.
176. Which of the following correctly expresses the difference between molar heat capacities \(C_p\) and \(C_v\)?
ⓐ. \(C_p – C_v = R\)
ⓑ. \(C_p = C_v\)
ⓒ. \(C_p – C_v = 2R\)
ⓓ. \(C_p – C_v = P\)
Correct Answer: \(C_p – C_v = R\)
Explanation: For an ideal gas, \(\Delta H = \Delta U + nR\Delta T\). Dividing by \(n\Delta T\) gives \(C_p – C_v = R\). This shows that enthalpy increases faster than internal energy with temperature, because \(C_p\) includes additional work of expansion at constant pressure, while \(C_v\) refers to heating at constant volume.
177. When one mole of ideal gas is heated at constant volume from \(T_1\) to \(T_2\), the relation between \(\Delta H\) and \(\Delta U\) is:
ⓐ. \(\Delta H = \Delta U + R(T_2 + T_1\)
ⓑ. \(\Delta H = \Delta U + P(T_2 – T_1\)
ⓒ. \(\Delta H = \Delta U + nR(T_2 – T_1\)
ⓓ. \(\Delta H = \Delta U + R(T_2 – T_1\)
Correct Answer: \(\Delta H = \Delta U + R(T_2 – T_1\)
Explanation: For one mole of an ideal gas, \(\Delta H – \Delta U = R\Delta T\). Since \(\Delta T = T_2 – T_1\), we get \(\Delta H = \Delta U + R(T_2 – T_1\). The term \(R\Delta T\) accounts for the additional expansion energy at constant pressure compared to heating at constant volume.
178. For a monatomic ideal gas, if \(C_v = \tfrac{3}{2}R\), what is \(C_p\)?
ⓐ. \(\tfrac{5}{2}R\)
ⓑ. \(\tfrac{3}{2}R\)
ⓒ. ( 2R )
ⓓ. \(\tfrac{7}{2}R\)
Correct Answer: \(\tfrac{5}{2}R\)
Explanation: For ideal gases, \(C_p = C_v + R\). Substituting \(C_v = \tfrac{3}{2}R\) gives \(C_p = \tfrac{3}{2}R + R = \tfrac{5}{2}R\). This relationship shows that at constant pressure, extra energy is needed not only to raise the temperature but also to perform expansion work against external pressure.
179. For the combustion of methane, \(CH_4(g) + 2O_2(g) → CO_2(g) + 2H_2O(l\), calculate \(\Delta H – \Delta U\) at 298 K.
ⓐ. ( -3RT )
ⓑ. ( 3RT )
ⓒ. ( 2RT )
ⓓ. ( -2RT )
Correct Answer: ( -2RT )
Explanation: The change in moles of gas \(\Delta n = (1) – (3) = -2\)? Wait, check: Products—\(CO_2(g) = 1\), reactants—\(CH_4 + 2O_2 = 3\), so \(\Delta n = 1 – 3 = -2\). The relation is \(\Delta H – \Delta U = \Delta nRT = -2RT\). Actually, for correct balancing including \(H_2O(l\) (liquid water, not gas), \(\Delta n = 1 – 3 = -2\). So the correct answer is ( -2RT ), which equals approximately –4.96 kJ per mole at 298 K.
180. For an ideal gas, which statement best describes the relation between ΔH and ΔU?
ⓐ. ΔH and ΔU are equal under all conditions
ⓑ. ΔH is always smaller than ΔU
ⓒ. ΔH = ΔU + nRΔT, showing dependence on temperature change
ⓓ. ΔH is independent of temperature
Correct Answer: ΔH = ΔU + nRΔT, showing dependence on temperature change
Explanation: For ideal gases, internal energy and enthalpy both depend only on temperature. The relationship \(\Delta H = \Delta U + nR\Delta T\) reflects how enthalpy includes both the internal energy change and the work of expansion at constant pressure. For isothermal processes \(\Delta T = 0\), both ΔH and ΔU become zero.
181. What is meant by heat capacity of a substance?
ⓐ. The amount of heat absorbed or released per mole per kelvin
ⓑ. The total energy content of a system
ⓒ. The quantity of heat required to raise the temperature of a substance by 1 K
ⓓ. The ratio of temperature to pressure of a system
Correct Answer: The quantity of heat required to raise the temperature of a substance by 1 K
Explanation: Heat capacity measures a substance’s ability to store thermal energy. It represents how much heat must be supplied to raise the temperature of the entire body by one kelvin (or one degree Celsius). The higher the heat capacity, the more energy required for the same temperature change. It depends on the substance’s mass, composition, and physical state.
182. What is the SI unit of heat capacity?
ⓐ. J mol⁻¹ K⁻¹
ⓑ. cal K⁻¹
ⓒ. J K⁻¹
ⓓ. W m⁻² K⁻¹
Correct Answer: J K⁻¹
Explanation: The SI unit of heat capacity is joule per kelvin (J K⁻¹). It expresses the amount of energy (in joules) needed to increase the temperature of a system by one kelvin. When heat capacity is measured per mole, the unit becomes J mol⁻¹ K⁻¹; when measured per gram, it is J g⁻¹ K⁻¹.
183. What is meant by molar heat capacity?
ⓐ. Heat capacity of one kilogram of substance
ⓑ. Heat capacity per mole of substance
ⓒ. Heat capacity per gram of substance
ⓓ. Heat capacity per litre of gas
Correct Answer: Heat capacity per mole of substance
Explanation: Molar heat capacity is defined as the quantity of heat required to raise the temperature of one mole of a substance by one kelvin. It is denoted by \(C_m\) and has units of J mol⁻¹ K⁻¹. It helps compare how different substances respond to heating on a per-mole basis, independent of sample size.
184. The relationship between heat ( q ), heat capacity ( C ), and temperature change \(\Delta T\) is:
ⓐ. \(q = C / \Delta T\)
ⓑ. \(q = C + \Delta T\)
ⓒ. \(q = C \times \Delta T\)
ⓓ. \(q = \Delta T / C\)
Correct Answer: \(q = C \times \Delta T\)
Explanation: The heat absorbed or released is proportional to the temperature change at constant heat capacity. The equation \(q = C \Delta T\) expresses this relationship. For example, if the heat capacity of a block is 200 J K⁻¹ and its temperature increases by 5 K, the heat absorbed is \(200 × 5 = 1000 \text{J}\).
185. If 500 J of heat raises the temperature of a metal block by 10 K, what is its heat capacity?
ⓐ. 10 J K⁻¹
ⓑ. 25 J K⁻¹
ⓒ. 50 J K⁻¹
ⓓ. 5000 J K⁻¹
Correct Answer: 50 J K⁻¹
Explanation: Using \(C = q / \Delta T\), \(C = 500 / 10 = 50 \text{J K}^{-1}\). This means 50 J of heat is required to raise the block’s temperature by one kelvin. The numerical value of (C) depends on the size and material of the block. Larger masses or materials with high specific heat have greater heat capacities.
186. What is the difference between specific heat capacity and molar heat capacity?
ⓐ. Specific heat is per gram; molar heat is per mole of substance
ⓑ. Specific heat is per mole; molar heat is per kilogram
ⓒ. Specific heat is per unit volume; molar heat is per gram
ⓓ. Both are identical quantities
Correct Answer: Specific heat is per gram; molar heat is per mole of substance
Explanation: Specific heat capacity ( c ) is the heat required to raise 1 g of a substance by 1 K, while molar heat capacity \(C_m\) refers to 1 mol. Their relationship is \(C_m = c × M\), where (M) is the molar mass. Both have the same physical meaning but differ in the quantity of matter considered.
187. What is the unit of molar heat capacity in SI system?
ⓐ. J g⁻¹ K⁻¹
ⓑ. J mol⁻¹ K⁻¹
ⓒ. cal K⁻¹
ⓓ. W K⁻¹
Correct Answer: J mol⁻¹ K⁻¹
Explanation: Molar heat capacity is expressed in joules per mole per kelvin (J mol⁻¹ K⁻¹). It quantifies how much heat is required to increase the temperature of one mole of a substance by one kelvin. This unit ensures direct comparison among different materials and allows use in gas equations like \(C_p – C_v = R\).
188. At constant volume, the molar heat capacity of an ideal gas is represented as:
ⓐ. \(C_p\)
ⓑ. \(C_v\)
ⓒ. \(C_m\)
ⓓ. \(C_t\)
Correct Answer: \(C_v\)
Explanation: The molar heat capacity measured at constant volume is \(C_v\). It represents the heat needed to raise the temperature of one mole of gas by 1 K while keeping volume constant, so no (P!V) work is done. Similarly, \(C_p\) is measured at constant pressure, where heat must also provide energy for expansion.
189. For an ideal gas, the relation between molar heat capacities \(C_p\) and \(C_v\) is:
ⓐ. \(C_p = C_v – R\)
ⓑ. \(C_p = C_v + R\)
ⓒ. \(C_p = R , C_v\)
ⓓ. \(C_p = C_v\)
Correct Answer: \(C_p = C_v + R\)
Explanation: The difference between \(C_p\) and \(C_v\) arises because at constant pressure, the system must perform expansion work. For an ideal gas, \(H = U + PV = U + nRT\). Differentiating gives \(dH = dU + nR\\,dT\), so per mole \(C_p – C_v = R\). This fundamental relation links enthalpy and internal-energy changes for gases.
190. If the molar heat capacity of a monatomic ideal gas at constant volume is \(\tfrac{3}{2}R\), what is its heat capacity at constant pressure?
ⓐ. ( 2R )
ⓑ. ( R )
ⓒ. \(\tfrac{5}{2}R\)
ⓓ. ( 3R )
Correct Answer: \(\tfrac{5}{2}R\)
Explanation: For any ideal gas, \(C_p – C_v = R\). Given \(C_v = \tfrac{3}{2}R\), we get \(C_p = \tfrac{3}{2}R + R = \tfrac{5}{2}R\). Physically, heating at constant pressure requires additional energy to perform (P!V) expansion work, so \(C_p\) always exceeds \(C_v\) by (R) for ideal gases.
191. What is the fundamental relation between molar heat capacities at constant pressure and volume for an ideal gas?
ⓐ. \(C_p = C_v – R\)
ⓑ. \(C_p = C_v / R\)
ⓒ. \(C_p = C_v \times R\)
ⓓ. \(C_p = C_v + R\)
Correct Answer: \(C_p = C_v + R\)
Explanation: For an ideal gas, \(H = U + PV = U + nRT\). Differentiating gives \(dH = dU + nR\\,dT\). Since \(C_p = \frac{dH}{dT}\) and \(C_v = \frac{dU}{dT}\), subtracting gives \(C_p – C_v = R\). This relation shows that at constant pressure, extra heat is required to do expansion work, making \(C_p\) always greater than \(C_v\).
192. Why is \(C_p\) greater than \(C_v\) for gases?
ⓐ. Because temperature increases more rapidly at constant pressure
ⓑ. Because at constant pressure, some heat is used to perform expansion work
ⓒ. Because pressure increases at constant volume
ⓓ. Because \(C_v\) includes potential energy only
Correct Answer: Because at constant pressure, some heat is used to perform expansion work
Explanation: At constant pressure, the system must expand when heated, doing \(P\Delta V\) work against the surroundings. Hence, more heat is required to raise the temperature by the same amount compared to constant volume heating. The difference between these two heat capacities equals the gas constant ( R ).
193. The equation \(C_p – C_v = R\) is valid for:
ⓐ. All solids and liquids
ⓑ. All real gases
ⓒ. Ideal gases only
ⓓ. Incompressible fluids
Correct Answer: Ideal gases only
Explanation: The derivation \(C_p – C_v = R\) assumes \(PV = nRT\) and that internal energy depends only on temperature — conditions valid for ideal gases. Real gases deviate from this relation because of intermolecular forces and non-ideal compressibility. For solids and liquids, the difference is much smaller and not equal to ( R ).
194. For 1 mole of an ideal gas, what is the numerical value of \(C_p – C_v\)?
ⓐ. 8.314 J mol⁻¹ K⁻¹
ⓑ. 4.18 J mol⁻¹ K⁻¹
ⓒ. 1.00 cal mol⁻¹ K⁻¹
ⓓ. 2.303 J mol⁻¹ K⁻¹
Correct Answer: 8.314 J mol⁻¹ K⁻¹
Explanation: The gas constant ( R ) has a value of 8.314 J mol⁻¹ K⁻¹. Therefore, for one mole of an ideal gas, \(C_p – C_v = R = 8.314 \text{ J mol}^{-1}\text{K}^{-1}\). This constant difference is crucial in thermodynamics and forms the basis for deriving the adiabatic relation \(PV^\gamma = \text{constant}\).
195. For a diatomic ideal gas, \(C_v = \tfrac{5}{2}R\). What is \(C_p\)?
ⓐ. \(\tfrac{3}{2}R\)
ⓑ. \(\tfrac{7}{2}R\)
ⓒ. ( 3R )
ⓓ. \(\tfrac{5}{2}R\)
Correct Answer: \(\tfrac{7}{2}R\)
Explanation: Using \(C_p – C_v = R\), if \(C_v = \tfrac{5}{2}R\), then \(C_p = \tfrac{5}{2}R + R = \tfrac{7}{2}R\). The higher value reflects the extra degrees of freedom (rotational modes) in diatomic molecules, which increase their energy storage and heat capacity.
196. If \(C_p = 29.1 , \text{J mol}^{-1}\text{K}^{-1}\) for air, what is \(C_v\)?
ⓐ. 29.1 J mol⁻¹ K⁻¹
ⓑ. 8.31 J mol⁻¹ K⁻¹
ⓒ. 25.0 J mol⁻¹ K⁻¹
ⓓ. 20.8 J mol⁻¹ K⁻¹
Correct Answer: 20.8 J mol⁻¹ K⁻¹
Explanation: For ideal gases, \(C_p – C_v = R = 8.314 , \text{J mol}^{-1}\text{K}^{-1}\). Therefore, \(C_v = C_p – R = 29.1 – 8.3 = 20.8 , \text{J mol}^{-1}\text{K}^{-1}\). This value corresponds closely to air’s behavior as a diatomic gas, whose theoretical \(C_v\) is \(\tfrac{5}{2}R \approx 20.8\).
197. What does the ratio \(\gamma = \dfrac{C_p}{C_v}\) represent?
ⓐ. Gas constant for the system
ⓑ. Specific heat ratio
ⓒ. Heat absorbed at constant pressure
ⓓ. Change in enthalpy per mole
Correct Answer: Specific heat ratio
Explanation: The ratio \(\gamma = \dfrac{C_p}{C_v}\) is called the adiabatic or specific heat ratio. It plays a central role in adiabatic processes and sound propagation. For monatomic gases, \(\gamma = 1.67\); for diatomic gases, \(\gamma \approx 1.4\). This ratio helps determine how gases behave when no heat exchange occurs.
198. Which equation links the ratio of heat capacities (γ) with adiabatic processes?
ⓐ. \(PV = nRT\)
ⓑ. \(PV^\gamma = \text{constant}\)
ⓒ. \(P/V = \gamma\)
ⓓ. \(T/P = \gamma\)
Correct Answer: \(PV^\gamma = \text{constant}\)
Explanation: For adiabatic reversible processes of an ideal gas, \(PV^\gamma = \text{constant}\). The exponent \(\gamma = \dfrac{C_p}{C_v}\) relates pressure and volume changes without heat exchange. This relation can be derived from the First Law using \(dq = 0\) and the ideal gas equation.
199. The value of \(\gamma\) for a monoatomic ideal gas is:
ⓐ. 1.00
ⓑ. 1.33
ⓒ. 1.67
ⓓ. 2.00
Correct Answer: 1.67
Explanation: For a monatomic ideal gas, \(C_v = \tfrac{3}{2}R\) and \(C_p = \tfrac{5}{2}R\). Thus, \(\gamma = \dfrac{C_p}{C_v} = \dfrac{5/2R}{3/2R} = \tfrac{5}{3} = 1.67\). This high value results from fewer degrees of freedom, which limits the energy modes available for storage, leading to greater temperature rise for the same heat input.
200. Which statement correctly explains why \(C_p – C_v = R\)?
ⓐ. Because pressure and volume are directly proportional
ⓑ. Because enthalpy and internal energy differ by ( nRT )
ⓒ. Because work done at constant volume equals ( nRT )
ⓓ. Because internal energy depends on both pressure and volume
Correct Answer: Because enthalpy and internal energy differ by ( nRT )
Explanation: Since \(H = U + PV\) and \(PV = nRT\) for an ideal gas, differentiating gives \(dH = dU + nR\\,dT\). Dividing by ( dT ) yields \(C_p – C_v = R\). This reflects that at constant pressure, heat must also perform expansion work equal to ( nR,dT ), hence the extra ( R ) term in \(C_p\) compared to \(C_v\).
You’re on Class 11 Chemistry MCQs – Chapter 6: Thermodynamics (Part 2). Based on the NCERT/CBSE syllabus, this chapter is crucial for Boards, JEE, NEET, and state-level exams. The full chapter offers 395 solved MCQs split into 4 parts for step-by-step mastery.
Focus of Part 2 (100 MCQs): First Law in action — q, w, ΔU, ΔH; expansion/compression work; ΔH = ΔU + ΔngRT (ideal gases); heat capacities (C, Cp, Cv); calorimetry and phase-change enthalpies; sign conventions and unit discipline. Perfect for building numerical speed + accuracy.
How to practice here
- Solve 10–20 MCQs at a time; check instant feedback and read the short explanation.
- Use the Heart on any MCQ to mark it as Favourite; then toggle the Favourite filter (near the Random button) to revise only your marked questions.
- Open Workspace under a question to jot quick reminders; notes auto-save and stay available for future sessions.
- Use Random to shuffle and test real understanding, not pattern memory.
- 👉 Total in chapter: 395 MCQs (100 + 100 + 100 + 95)
- 👉 This page: Second set of 100 solved MCQs
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