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201. Which principle is the basis of calorimetry?
ⓐ. Conservation of mass
ⓑ. Conservation of energy
ⓒ. Conservation of momentum
ⓓ. Conservation of charge
Correct Answer: Conservation of energy
Explanation: Calorimetry is based on the conservation of energy, which states that heat lost by a hot body equals heat gained by a cold body, assuming negligible heat exchange with surroundings.
202. A 100 g piece of metal at $150^\circ C$ is dropped into 200 g of water at $25^\circ C$. Final temperature is $30^\circ C$. Calculate the specific heat of the metal. (Specific heat of water = $4.18 \, J/gK$)
ⓐ. $0.20 \, J/gK$
ⓑ. $0.25 \, J/gK$
ⓒ. $0.30 \, J/gK$
ⓓ. $0.35 \, J/gK$
Correct Answer: $0.25 \, J/gK$
Explanation: Heat lost by metal = Heat gained by water$$100c(150 – 30) = 200 \times 4.18 (30 – 25).$$
$$100c \times 120 = 200 \times 20.9 = 4180.$$
$$12000c = 4180 \Rightarrow c \approx 0.25 \, J/gK.$$
203. In calorimetry experiments, why is a calorimeter usually made of copper?
ⓐ. Because copper is cheap
ⓑ. Because copper has very low density
ⓒ. Because copper has high thermal conductivity and low specific heat
ⓓ. Because copper has high resistance to corrosion
Correct Answer: Because copper has high thermal conductivity and low specific heat
Explanation: Copper allows rapid transfer of heat between substances and has low heat capacity, so it does not absorb much heat itself, minimizing errors in measurement.
204. A calorimeter of mass 150 g contains 200 g of water at $30^\circ C$. A 100 g hot solid is dropped into it, and equilibrium is reached at $40^\circ C$. If the solid was initially at $100^\circ C$ and specific heat of calorimeter copper = $0.39 \, J/gK$, calculate the specific heat of the solid. (Specific heat of water = $4.18 \, J/gK$)
ⓐ. $0.60 \, J/gK$
ⓑ. $0.70 \, J/gK$
ⓒ. $0.80 \, J/gK$
ⓓ. $0.90 \, J/gK$
Correct Answer: $0.70 \, J/gK$
Explanation: Heat lost by solid = Heat gained by (water + calorimeter).$$100c(100 – 40) = (200 \times 4.18 + 150 \times 0.39)(40 – 30).$$
$$100c \times 60 = (836 + 58.5) \times 10 = 8945.$$
$$6000c = 8945 \Rightarrow c \approx 0.70 \, J/gK.$$
205. Which of the following is an application of calorimetry in real life?
ⓐ. Determining speed of sound
ⓑ. Determining the specific heat of unknown substances
ⓒ. Measuring electric resistance
ⓓ. Calculating gravitational force
Correct Answer: Determining the specific heat of unknown substances
Explanation: Calorimetry is used to determine specific heats, latent heats, and heats of reactions by applying the principle of energy conservation.
206. A 50 g ice cube at $0^\circ C$ is dropped into 200 g of water at $30^\circ C$. Find the final temperature after thermal equilibrium is reached. (Latent heat of fusion of ice = $334 \, J/g$, specific heat of water = $4.18 \, J/gK$).
ⓐ. $0^\circ C$
ⓑ. $5^\circ C$
ⓒ. $10^\circ C$
ⓓ. $15^\circ C$
Correct Answer: $5^\circ C$
Explanation: Heat required to melt ice = $50 \times 334 = 16700 J$.
Heat lost by water cooling from 30 °C to $T$: $200 \times 4.18 (30 – T)$.$$200 \times 4.18 (30 – T) = 16700 + 50 \times 4.18 (T – 0).$$
Solve gives $T \approx 5^\circ C$.
207. In heat transfer by conduction, thermal conductivity of water compared to metals is:
ⓐ. Much higher
ⓑ. Much lower
ⓒ. Exactly the same
ⓓ. Infinite
Correct Answer: Much lower
Explanation: Metals such as copper and aluminum are good conductors with high $k$, while water has a much lower thermal conductivity, so it is used as a coolant in some applications.
208. A calorimeter contains 250 g of water at $20^\circ C$. A 100 g piece of metal at $80^\circ C$ is placed in it. Final equilibrium temperature is $25^\circ C$. Find the specific heat of the metal. (Specific heat of water = $4.18 \, J/gK$).
ⓐ. $0.35 \, J/gK$
ⓑ. $0.45 \, J/gK$
ⓒ. $0.55 \, J/gK$
ⓓ. $0.65 \, J/gK$
Correct Answer: $0.35 \, J/gK$
Explanation: Heat lost by metal = Heat gained by water.$$100c(80 – 25) = 250 \times 4.18 (25 – 20).$$
$$100c \times 55 = 250 \times 20.9 = 5225.$$
$$5500c = 5225 \Rightarrow c \approx 0.35 \, J/gK.$$
209. Which heat transfer process is primarily responsible for sea and land breezes?
ⓐ. Conduction
ⓑ. Convection
ⓒ. Radiation
ⓓ. Evaporation
Correct Answer: Convection
Explanation: Sea and land breezes occur because air above land and sea is heated unevenly, creating density differences. The bulk movement of air (a fluid) is convection.
210. A 100 g piece of steam at $100^\circ C$ is passed into 500 g of water at $20^\circ C$. Find the final temperature when equilibrium is reached. (Latent heat of vaporization = $2260 \, J/g$, $c_{water} = 4.18 \, J/gK$).
ⓐ. $60^\circ C$
ⓑ. $70^\circ C$
ⓒ. $80^\circ C$
ⓓ. $90^\circ C$
Correct Answer: $80^\circ C$
Explanation: Steam condenses and releases heat = $100 \times 2260 = 226000 J$. This plus cooling of condensed water raises the 500 g water temperature. Solving energy balance gives final temperature close to $80^\circ C$.
211. Which of the following is a state variable in thermodynamics?
ⓐ. Work
ⓑ. Heat
ⓒ. Pressure
ⓓ. Path followed by a process
Correct Answer: Pressure
Explanation: State variables depend only on the current state of the system, not on how the state was reached. Pressure, volume, temperature, and internal energy are state variables. Work and heat are path functions, not state variables.
212. Which among the following is not a state variable?
ⓐ. Temperature
ⓑ. Internal energy
ⓒ. Volume
ⓓ. Heat
Correct Answer: Heat
Explanation: Heat is energy in transit, not a property stored in the system. State variables like temperature, internal energy, and volume describe the system’s equilibrium state, while heat depends on the process.
213. A gas initially at pressure $2 \, atm$, volume $3 \, L$, and temperature $300 \, K$. If it changes to pressure $1 \, atm$, volume $6 \, L$, temperature $300 \, K$, which of the following remains unchanged?
ⓐ. Pressure
ⓑ. Volume
ⓒ. Temperature
ⓓ. Number of molecules
Correct Answer: Temperature
Explanation: In this process, pressure and volume change, but the temperature remains constant. Since temperature is a state variable, it fully describes the thermal state here.
214. The SI unit of pressure is:
ⓐ. Dyne/cm$^2$
ⓑ. Newton
ⓒ. Pascal
ⓓ. Joule
Correct Answer: Pascal
Explanation: 1 Pascal (Pa) = 1 Newton per square meter ($1 \, N/m^2$). It is the SI unit of pressure, while other units (dyne/cm², atm) are non-SI but used in practice.
215. Which of the following is a correct pair of conjugate state variables?
ⓐ. Pressure and Volume
ⓑ. Heat and Work
ⓒ. Entropy and Path
ⓓ. Time and Temperature
Correct Answer: Pressure and Volume
Explanation: Pressure (intensive) and volume (extensive) are conjugate variables in thermodynamics. Their product often appears in work expressions ($PdV$). Heat and work are not state functions, while entropy and path are unrelated.
216. If a gas at pressure $2 \times 10^5 \, Pa$ occupies volume $0.01 \, m^3$, calculate the work done when volume increases to $0.03 \, m^3$ at constant pressure.
ⓐ. $1000 \, J$
ⓑ. $2000 \, J$
ⓒ. $3000 \, J$
ⓓ. $4000 \, J$
Correct Answer: $2000 \, J$
Explanation: Work done $W = P \Delta V = 2 \times 10^5 (0.03 – 0.01) = 2 \times 10^5 \times 0.02 = 4000 J$.
Correction: That equals 4000 J, so correct option is D.
217. Temperature is defined as:
ⓐ. Measure of total internal energy of a system
ⓑ. Measure of average kinetic energy of particles in a system
ⓒ. Amount of heat supplied to a body
ⓓ. Pressure exerted by a gas on container walls
Correct Answer: Measure of average kinetic energy of particles in a system
Explanation: In kinetic theory, temperature corresponds to the average translational kinetic energy of molecules. Internal energy is related but includes other contributions like rotational and vibrational energies.
218. A cylinder contains a gas at volume $2 \, L$, pressure $4 \, atm$. If the gas is compressed to $1 \, L$ at constant temperature, what is the final pressure?
ⓐ. $2 \, atm$
ⓑ. $4 \, atm$
ⓒ. $6 \, atm$
ⓓ. $8 \, atm$
Correct Answer: $8 \, atm$
Explanation: By Boyle’s law ($P_1V_1 = P_2V_2$):$$4 \times 2 = P_2 \times 1 \Rightarrow P_2 = 8 \, atm.$$
219. Which of the following is an intensive property among state variables?
ⓐ. Pressure
ⓑ. Volume
ⓒ. Internal energy
ⓓ. Enthalpy
Correct Answer: Pressure
Explanation: Intensive properties do not depend on the size or amount of the system (e.g., temperature, pressure, density). Extensive properties (e.g., volume, enthalpy, internal energy) depend on the amount of matter.
220. A gas at $27^\circ C$ has pressure $1 \, atm$. If temperature is increased to $327^\circ C$ at constant volume, what will be the final pressure?
ⓐ. $2 \, atm$
ⓑ. $3 \, atm$
ⓒ. $4 \, atm$
ⓓ. $5 \, atm$
Correct Answer: $2 \, atm$
Explanation: At constant volume, $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.$$\frac{1}{300} = \frac{P_2}{600} \Rightarrow P_2 = 2 \, atm.$$
221. The ideal gas law is expressed as:
ⓐ. $PV = nRT$
ⓑ. $PV = RT$
ⓒ. $P = \frac{n}{RT}$
ⓓ. $V = \frac{P}{nR}$
Correct Answer: $PV = nRT$
Explanation: The ideal gas law combines Boyle’s, Charles’, and Avogadro’s laws. Here, $P$ = pressure, $V$ = volume, $n$ = number of moles, $R$ = gas constant, and $T$ = absolute temperature.
222. One mole of an ideal gas at STP occupies:
ⓐ. $11.2 \, L$
ⓑ. $22.4 \, L$
ⓒ. $33.6 \, L$
ⓓ. $44.8 \, L$
Correct Answer: $22.4 \, L$
Explanation: At STP (1 atm, 273 K), molar volume of an ideal gas is $22.4 \, L$. This follows directly from the ideal gas equation $PV = nRT$.
223. The gas constant $R$ has the value:
ⓐ. $8.314 \, J/mol \, K$
ⓑ. $4.18 \, J/g \, K$
ⓒ. $1.013 \times 10^5 \, Pa$
ⓓ. $6.022 \times 10^{23} \, J$
Correct Answer: $8.314 \, J/mol \, K$
Explanation: Universal gas constant $R = 8.314 \, J/mol \, K$. Option B is specific heat of water, option C is atmospheric pressure, and option D is Avogadro’s number times Joule (not correct).
224. A sample of gas has volume $5 \, L$, pressure $2 \, atm$, temperature $300 \, K$. Calculate the number of moles. (Take $R = 0.0821 \, L\, atm/mol\, K$).
ⓐ. 0.20 mol
ⓑ. 0.30 mol
ⓒ. 0.40 mol
ⓓ. 0.50 mol
Correct Answer: 0.40 mol
Explanation: From ideal gas law,$$n = \frac{PV}{RT} = \frac{2 \times 5}{0.0821 \times 300}.$$
$$n = \frac{10}{24.63} \approx 0.40 \, mol.$$
225. The Van der Waals equation for real gases is:
ⓐ. $(P + \frac{a}{V^2})(V – b) = RT$
ⓑ. $(P – \frac{a}{V^2})(V + b) = RT$
ⓒ. $(P + \frac{a}{V^2})(V + b) = RT$
ⓓ. $(P – \frac{a}{V^2})(V – b) = RT$
Correct Answer: $(P + \frac{a}{V^2})(V – b) = RT$
Explanation: Van der Waals corrected the ideal gas law for real gases by adding:
– Pressure correction: $P + \frac{a}{V^2}$ (accounts for intermolecular attractions),
– Volume correction: $V – b$ (accounts for finite molecular size).
226. The constant $a$ in Van der Waals equation accounts for:
ⓐ. Volume of molecules
ⓑ. Intermolecular attractions
ⓒ. Kinetic energy of molecules
ⓓ. Temperature of gas
Correct Answer: Intermolecular attractions
Explanation: $a$ corrects for pressure reduction due to molecular attractions. Larger $a$ indicates stronger intermolecular forces (e.g., NH₃ > H₂).
227. The constant $b$ in Van der Waals equation accounts for:
ⓐ. Pressure correction
ⓑ. Molecular volume (excluded volume)
ⓒ. Energy of collisions
ⓓ. Specific heat capacity
Correct Answer: Molecular volume (excluded volume)
Explanation: $b$ is the excluded volume per mole, representing the finite size of molecules. The effective free volume available for molecular motion is reduced to $(V – b)$.
228. A real gas behaves most like an ideal gas under conditions of:
ⓐ. Low temperature and high pressure
ⓑ. High temperature and low pressure
ⓒ. Low temperature and low pressure
ⓓ. High temperature and high pressure
Correct Answer: High temperature and low pressure
Explanation: At high temperature, intermolecular forces become negligible; at low pressure, the volume of molecules compared to container volume is negligible. Hence, gas approximates ideal behavior.
229. Calculate the pressure of 1 mole of gas at 300 K in 10 L using Van der Waals equation, given $a = 1.36 \, atm \, L^2/mol^2$, $b = 0.031 \, L/mol$, $R = 0.0821 \, L \, atm/mol \, K$.
ⓐ. 2.40 atm
ⓑ. 2.46 atm
ⓒ. 2.52 atm
ⓓ. 2.60 atm
Correct Answer: 2.46 atm
Explanation: $$P = \frac{nRT}{V – nb} – \frac{a n^2}{V^2}.$$
$$= \frac{1 \times 0.0821 \times 300}{10 – 0.031} – \frac{1.36 \times 1^2}{100}.$$
$$= \frac{24.63}{9.969} – 0.0136 = 2.47 – 0.0136 \approx 2.46 \, atm.$$
230. Which gas deviates least from ideal behavior?
ⓐ. H₂
ⓑ. CO₂
ⓒ. NH₃
ⓓ. SO₂
Correct Answer: H₂
Explanation: Hydrogen (H₂) molecules are very small with weak intermolecular forces, hence deviations from ideal gas law are minimal. Gases like NH₃ and SO₂ deviate more due to strong intermolecular attractions.
231. Why are equations of state like the ideal gas law important in thermodynamic analysis?
ⓐ. They help calculate molecular masses directly
ⓑ. They provide a relation between measurable macroscopic variables (P, V, T)
ⓒ. They describe microscopic structure of atoms
ⓓ. They replace the First Law of Thermodynamics
Correct Answer: They provide a relation between measurable macroscopic variables (P, V, T)
Explanation: Equations of state, such as $PV = nRT$, connect measurable quantities like pressure, volume, and temperature. This allows engineers and scientists to model real systems without needing molecular-level details.
232. Why is the Van der Waals equation used in modeling real gases instead of the ideal gas law?
ⓐ. It ignores pressure and volume effects
ⓑ. It accounts for molecular size and intermolecular forces
ⓒ. It eliminates the need for temperature measurement
ⓓ. It applies only at absolute zero
Correct Answer: It accounts for molecular size and intermolecular forces
Explanation: Real gases deviate from ideal behavior at high pressure and low temperature. The Van der Waals corrections ($a, b$) account for attractions and finite molecular size, making the model more accurate.
233. In thermodynamic modeling, why is it important to identify state variables (P, V, T, U, H)?
ⓐ. They are always path functions
ⓑ. They uniquely define the state of the system
ⓒ. They can be ignored for cyclic processes
ⓓ. They are only valid for solids
Correct Answer: They uniquely define the state of the system
Explanation: State variables describe the equilibrium state of a system. Knowledge of a sufficient set of these variables allows complete thermodynamic modeling and prediction of system behavior.
234. Which of the following is an application of equations of state in engineering?
ⓐ. Designing heat engines and refrigerators
ⓑ. Measuring electrical resistance
ⓒ. Studying light refraction in optics
ⓓ. Analyzing sound wave frequencies
Correct Answer: Designing heat engines and refrigerators
Explanation: Equations of state are used in analyzing thermodynamic cycles like Otto, Diesel, Rankine, and refrigeration cycles. They allow calculation of work output, efficiency, and heat transfer.
235. A cylinder contains 2 moles of an ideal gas at 300 K. Using the ideal gas law, calculate the pressure if the volume is 5 L. (Take $R = 0.0821 \, L \, atm/mol \, K$).
ⓐ. 8 atm
ⓑ. 9 atm
ⓒ. 10 atm
ⓓ. 12 atm
Correct Answer: 10 atm
Explanation: $$P = \frac{nRT}{V} = \frac{2 \times 0.0821 \times 300}{5}.$$
$$= \frac{49.26}{5} \approx 9.85 \, atm \approx 10 \, atm.$$
236. Why is thermodynamic modeling essential in chemical engineering?
ⓐ. To determine atomic numbers of elements
ⓑ. To optimize processes like distillation, combustion, and refrigeration
ⓒ. To measure nuclear half-life
ⓓ. To replace experimental results
Correct Answer: To optimize processes like distillation, combustion, and refrigeration
Explanation: Thermodynamic models predict heat, work, and phase behavior of substances, which are vital for process design and optimization in chemical engineering.
237. Which of the following is a limitation of the ideal gas law in modeling real gases?
ⓐ. It assumes zero molecular interactions
ⓑ. It requires very high computational power
ⓒ. It cannot be applied to helium
ⓓ. It ignores conservation of energy
Correct Answer: It assumes zero molecular interactions
Explanation: The ideal gas law assumes point particles with no intermolecular forces. This fails at high pressure and low temperature where real gases show significant interactions.
238. A Van der Waals gas has constants $a = 1.36 \, atm \, L^2/mol^2$, $b = 0.031 \, L/mol$. For 1 mole of gas at 300 K and volume 10 L, calculate pressure using Van der Waals equation. (Take $R = 0.0821 \, L\, atm/mol\, K$).
ⓐ. 2.30 atm
ⓑ. 2.40 atm
ⓒ. 2.50 atm
ⓓ. 2.60 atm
Correct Answer: 2.40 atm
Explanation: $$P = \frac{nRT}{V – nb} – \frac{a n^2}{V^2}.$$
$$= \frac{1 \times 0.0821 \times 300}{10 – 0.031} – \frac{1.36}{100}.$$
$$= \frac{24.63}{9.969} – 0.0136 = 2.47 – 0.0136 = 2.46 \, atm \approx 2.40 \, atm.$$
239. In thermodynamic modeling, why are intensive properties (P, T, density) more useful than extensive properties?
ⓐ. They depend on system size
ⓑ. They remain independent of system size, allowing generalized equations
ⓒ. They are difficult to measure
ⓓ. They are only valid for liquids
Correct Answer: They remain independent of system size, allowing generalized equations
Explanation: Intensive properties like pressure and temperature do not depend on the amount of matter, making them universally applicable in equations of state and thermodynamic models.
240. Why is the ideal gas law often used in preliminary thermodynamic analysis instead of the Van der Waals equation?
ⓐ. It is less accurate
ⓑ. It is simpler and sufficiently accurate at normal conditions
ⓒ. It does not require temperature measurement
ⓓ. It ignores energy conservation
Correct Answer: It is simpler and sufficiently accurate at normal conditions
Explanation: For gases at moderate pressure and high temperature, deviations from ideal behavior are negligible. The ideal gas law provides a quick and reasonably accurate model for analysis, while Van der Waals is used when accuracy is critical.
241. In an isothermal process, which parameter remains constant?
ⓐ. Pressure
ⓑ. Volume
ⓒ. Temperature
ⓓ. Entropy
Correct Answer: Temperature
Explanation: In an isothermal process, temperature remains constant. For an ideal gas, internal energy depends only on temperature, so $\Delta U = 0$, which means $Q = W$.
242. Which of the following is true for an adiabatic process?
ⓐ. Heat exchange with surroundings is zero
ⓑ. Work done is zero
ⓒ. Temperature remains constant
ⓓ. Pressure remains constant
Correct Answer: Heat exchange with surroundings is zero
Explanation: In an adiabatic process, no heat is transferred ($Q = 0$). Any change in internal energy comes entirely from work done on or by the system: $\Delta U = -W$.
243. In an isobaric process, the work done is calculated as:
ⓐ. $W = nRT\ln{\frac{V_2}{V_1}}$
ⓑ. $W = P\Delta V$
ⓒ. $W = 0$
ⓓ. $W = \Delta U$
Correct Answer: $W = P\Delta V$
Explanation: At constant pressure, the work done equals the pressure multiplied by the change in volume. This is common in heating or cooling processes at constant atmospheric pressure.
244. In an isochoric process, the work done is:
ⓐ. Zero
ⓑ. Equal to heat supplied
ⓒ. Equal to internal energy change
ⓓ. Equal to enthalpy change
Correct Answer: Zero
Explanation: In an isochoric process, volume remains constant ($\Delta V = 0$). Hence, work done ($W = P \Delta V$) is zero. Any heat supplied only changes internal energy.
245. A gas expands isothermally at 300 K from 2 L to 6 L. Calculate the work done by 1 mole of gas. (Take $R = 8.314 \, J/molK$).
ⓐ. $2740 \, J$
ⓑ. $2990 \, J$
ⓒ. $3300 \, J$
ⓓ. $3600 \, J$
Correct Answer: $2990 \, J$
Explanation: For isothermal process:$$W = nRT \ln{\frac{V_2}{V_1}} = 1 \times 8.314 \times 300 \times \ln(3).$$
$\ln(3) \approx 1.099$.$$W \approx 2494 \times 1.099 = 2990 \, J.$$
246. In an adiabatic process for an ideal gas, the relation between pressure and volume is:
ⓐ. $PV = \text{constant}$
ⓑ. $P + V = \text{constant}$
ⓒ. $PV^\gamma = \text{constant}$
ⓓ. $P/V = \text{constant}$
Correct Answer: $PV^\gamma = \text{constant}$
Explanation: For adiabatic processes, $PV^\gamma = \text{constant}$, where $\gamma = C_p/C_v$. This relation is derived from the First Law and the condition $Q = 0$.
247. A gas undergoes isochoric heating where its temperature increases by 50 K. Which statement is correct?
ⓐ. Work done is zero, internal energy increases
ⓑ. Work done is zero, internal energy constant
ⓒ. Work done equals heat supplied
ⓓ. Work done equals enthalpy change
Correct Answer: Work done is zero, internal energy increases
Explanation: Isochoric processes involve no volume change, so $W = 0$. The heat supplied increases the internal energy directly, since $\Delta U = Q$.
248. Which process is represented by a horizontal line in a PV diagram?
ⓐ. Isothermal
ⓑ. Isochoric
ⓒ. Isobaric
ⓓ. Adiabatic
Correct Answer: Isobaric
Explanation: In a PV diagram, a horizontal line indicates constant pressure while volume changes. Hence, it represents an isobaric process.
249. A system undergoes adiabatic compression. What happens to its temperature?
ⓐ. Decreases
ⓑ. Increases
ⓒ. Remains constant
ⓓ. Becomes zero
Correct Answer: Increases
Explanation: In adiabatic compression, work is done on the gas, which increases internal energy, leading to an increase in temperature. Conversely, in adiabatic expansion, temperature decreases.
250. Which of the following is correct for an isothermal process involving an ideal gas?
ⓐ. $\Delta U = Q$
ⓑ. $\Delta U = 0$
ⓒ. $W = 0$
ⓓ. $Q = 0$
Correct Answer: $\Delta U = 0$
Explanation: For an ideal gas in an isothermal process, internal energy depends only on temperature. Since temperature remains constant, $\Delta U = 0$. Hence, $Q = W$.
251. In a PV diagram, the area under the curve represents:
ⓐ. Change in temperature
ⓑ. Change in internal energy
ⓒ. Work done by the system
ⓓ. Heat transferred
Correct Answer: Work done by the system
Explanation: The work done by or on a gas during a thermodynamic process is equal to the area under the $P-V$ curve. Heat transfer depends on the process, but only work is represented geometrically.
252. An isothermal process on a PV diagram appears as:
ⓐ. A vertical line
ⓑ. A horizontal line
ⓒ. A rectangular hyperbola
ⓓ. A straight diagonal line
Correct Answer: A rectangular hyperbola
Explanation: For isothermal processes, $PV = \text{constant}$. Thus, pressure decreases hyperbolically with volume, forming a rectangular hyperbola in the PV diagram.
253. Which process is represented by a vertical line on a PV diagram?
ⓐ. Isothermal
ⓑ. Isochoric
ⓒ. Isobaric
ⓓ. Adiabatic
Correct Answer: Isochoric
Explanation: In isochoric processes, volume remains constant, so the PV diagram is a vertical line with changing pressure.
254. An isobaric process on a PV diagram is represented by:
ⓐ. A horizontal line
ⓑ. A vertical line
ⓒ. A hyperbola
ⓓ. A straight diagonal
Correct Answer: A horizontal line
Explanation: At constant pressure, $P$ does not change while $V$ changes, so the PV diagram shows a horizontal line.
255. In an adiabatic process, the PV curve is:
ⓐ. Flatter than isothermal
ⓑ. Steeper than isothermal
ⓒ. Same as isothermal
ⓓ. A straight vertical line
Correct Answer: Steeper than isothermal
Explanation: For an adiabatic process, $PV^\gamma = \text{constant}$. Since $\gamma > 1$, the PV curve is steeper compared to an isothermal hyperbola.
256. In a TS diagram, the area under the curve represents:
ⓐ. Work done
ⓑ. Heat transferred
ⓒ. Change in entropy
ⓓ. Change in pressure
Correct Answer: Heat transferred
Explanation: In a $T-S$ diagram, the area under the curve corresponds to the heat transfer $Q = \int T \, dS$. This makes the TS diagram useful for analyzing cycles like Carnot and Rankine.
257. An isothermal process on a TS diagram appears as:
ⓐ. A horizontal line
ⓑ. A vertical line
ⓒ. A curved line
ⓓ. A diagonal line
Correct Answer: A horizontal line
Explanation: In an isothermal process, temperature $T$ is constant, so the TS diagram is a horizontal line at fixed $T$, with entropy $S$ changing.
258. Which process appears as a vertical line on a TS diagram?
ⓐ. Isothermal
ⓑ. Isochoric
ⓒ. Adiabatic (reversible)
ⓓ. Isobaric
Correct Answer: Adiabatic (reversible)
Explanation: For a reversible adiabatic process, $Q = 0$, so entropy $S$ remains constant. Hence, on a TS diagram, it is a vertical line.
259. In a TS diagram, the cycle efficiency can be calculated as:
ⓐ. Ratio of total work to heat input
ⓑ. Ratio of area enclosed by cycle to heat input
ⓒ. Area enclosed by the cycle divided by total heat transfer
ⓓ. Area enclosed by the cycle divided by input temperature
Correct Answer: Area enclosed by the cycle divided by total heat transfer
Explanation: The area enclosed in a TS diagram represents net heat converted to work. Thus, efficiency can be interpreted as the ratio of useful work (area of cycle) to heat absorbed.
260. On a PV diagram, which cycle appears as a rectangle?
ⓐ. Otto cycle
ⓑ. Carnot cycle
ⓒ. Stirling cycle
ⓓ. Diesel cycle
Correct Answer: Stirling cycle
Explanation: The Stirling cycle has two isothermal processes and two isochoric processes. On a PV diagram, these combine to form a rectangular shape.
261. In an isothermal expansion of an ideal gas, the work done is given by:
ⓐ. $W = P \Delta V$
ⓑ. $W = nRT \ln \frac{V_2}{V_1}$
ⓒ. $W = \Delta U$
ⓓ. $W = 0$
Correct Answer: $W = nRT \ln \frac{V_2}{V_1}$
Explanation: For an isothermal process, $T$ is constant. Since internal energy depends only on $T$, $\Delta U = 0$. By the First Law, $Q = W$. Integration of $PdV$ with $P = \frac{nRT}{V}$ gives $W = nRT \ln \frac{V_2}{V_1}$.
262. A mole of ideal gas expands isothermally at 300 K from 5 L to 10 L. Calculate the work done. (Take $R = 8.314 \, J/molK$).
ⓐ. $1700 \, J$
ⓑ. $1725 \, J$
ⓒ. $1730 \, J$
ⓓ. $1740 \, J$
Correct Answer: $1725 \, J$
Explanation: $$W = nRT \ln \frac{V_2}{V_1} = 1 \times 8.314 \times 300 \times \ln 2.$$
$$= 2494.2 \times 0.693 = 1728 \, J \approx 1730 J.$$
263. In an adiabatic process, heat exchange $Q$ is:
ⓐ. Equal to work done
ⓑ. Zero
ⓒ. Equal to internal energy change
ⓓ. Infinite
Correct Answer: Zero
Explanation: By definition, an adiabatic process is thermally insulated, so no heat is exchanged ($Q = 0$). Any work done changes the internal energy: $\Delta U = -W$.
264. A gas expands adiabatically and does $250 \, J$ of work. What is the change in internal energy?
ⓐ. $+250 \, J$
ⓑ. $-250 \, J$
ⓒ. $0 \, J$
ⓓ. $+500 \, J$
Correct Answer: $-250 \, J$
Explanation: For adiabatic process, $Q = 0$. By First Law:$$\Delta U = Q – W = 0 – 250 = -250 \, J.$$
265. In an isobaric process, heat supplied is related to change in enthalpy as:
ⓐ. $Q = \Delta U$
ⓑ. $Q = \Delta H$
ⓒ. $Q = W$
ⓓ. $Q = 0$
Correct Answer: $Q = \Delta H$
Explanation: At constant pressure, heat supplied equals change in enthalpy: $Q_p = \Delta H$. This is a key relation for heating processes in open systems.
266. A gas at constant pressure of $2 \times 10^5 \, Pa$ expands from 0.01 m³ to 0.03 m³. Calculate the work done.
ⓐ. $2000 \, J$
ⓑ. $3000 \, J$
ⓒ. $4000 \, J$
ⓓ. $5000 \, J$
Correct Answer: $4000 \, J$
Explanation: $$W = P \Delta V = 2 \times 10^5 (0.03 – 0.01).$$
$$W = 2 \times 10^5 \times 0.02 = 4000 \, J.$$
267. In an isochoric process, the heat supplied is:
ⓐ. Converted entirely into work
ⓑ. Equal to change in internal energy
ⓒ. Always zero
ⓓ. Equal to enthalpy change
Correct Answer: Equal to change in internal energy
Explanation: In an isochoric process ($\Delta V = 0$), no work is done ($W = 0$). From the First Law, $Q = \Delta U$.
268. A 2 mole ideal gas is heated at constant volume from 300 K to 400 K. Find the heat supplied. (For a monoatomic gas, $C_v = \frac{3}{2}R$, $R = 8.314 \, J/molK$).
ⓐ. $2500 \, J$
ⓑ. $3741 \, J$
ⓒ. $4988 \, J$
ⓓ. $6235 \, J$
Correct Answer: $4988 \, J$
Explanation: $$Q = n C_v \Delta T = 2 \times \frac{3}{2} \times 8.314 \times (400 – 300).$$
$$Q = 3 \times 8.314 \times 100 = 2494 \times 2 = 4988 \, J.$$
269. For a reversible adiabatic process of an ideal gas, the relation between temperature and volume is:
ⓐ. $TV = \text{constant}$
ⓑ. $T/V = \text{constant}$
ⓒ. $TV^{\gamma-1} = \text{constant}$
ⓓ. $T^\gamma V = \text{constant}$
Correct Answer: $TV^{\gamma-1} = \text{constant}$
Explanation: The relation for adiabatic process is $PV^\gamma = \text{constant}$. Using ideal gas law, this transforms to $TV^{\gamma-1} = \text{constant}$.
270. A gas undergoes an isothermal expansion from 2 L to 8 L at 300 K. Calculate the heat absorbed if 1 mole of the gas is used. (Take $R = 8.314 \, J/molK$).
ⓐ. $3450 \, J$
ⓑ. $4600 \, J$
ⓒ. $5170 \, J$
ⓓ. $6900 \, J$
Correct Answer: $5170 \, J$
Explanation: For isothermal process, $Q = W$.$$Q = nRT \ln \frac{V_2}{V_1} = 1 \times 8.314 \times 300 \times \ln(4).$$
$\ln(4) = 1.386$.$$Q = 2494.2 \times 1.386 \approx 5170 \, J.$$
271. The basic principle of a heat engine is:
ⓐ. To convert all heat energy into work
ⓑ. To transfer heat from a cold body to a hot body without work
ⓒ. To convert part of heat absorbed from a hot reservoir into work
ⓓ. To convert mechanical work fully into heat
Correct Answer: To convert part of heat absorbed from a hot reservoir into work
Explanation: A heat engine operates between a hot reservoir and a cold reservoir. It absorbs heat $Q_H$ from the source, converts part of it into work $W$, and rejects the remaining $Q_C$ to the sink.
272. Which of the following is NOT a component of a heat engine?
ⓐ. Source (hot reservoir)
ⓑ. Sink (cold reservoir)
ⓒ. Working substance
ⓓ. Transformer
Correct Answer: Transformer
Explanation: A heat engine requires three main components: hot source, working substance (e.g., gas, steam), and cold sink. A transformer is an electrical device, not part of a heat engine.
273. The working substance in a heat engine is:
ⓐ. The material that transfers heat directly between reservoirs
ⓑ. The medium that absorbs heat and performs work
ⓒ. Always air only
ⓓ. Always water only
Correct Answer: The medium that absorbs heat and performs work
Explanation: The working substance (e.g., steam in steam engines, air in gas turbines) is the system that absorbs heat from the source, expands to perform work, and rejects heat to the sink.
274. In a steam engine, the boiler acts as:
ⓐ. The sink
ⓑ. The source
ⓒ. The working substance
ⓓ. The piston
Correct Answer: The source
Explanation: The boiler supplies heat to the working substance (steam). It serves as the hot reservoir or source in the thermodynamic cycle.
275. The function of the condenser in a steam engine is:
ⓐ. To supply heat to the working substance
ⓑ. To convert work into heat
ⓒ. To serve as a cold reservoir (sink)
ⓓ. To compress steam to high pressure
Correct Answer: To serve as a cold reservoir (sink)
Explanation: The condenser allows steam to release heat to the environment, condensing into liquid. This maintains the cycle by providing a cold sink for heat rejection.
276. Which of the following represents the efficiency of a heat engine?
ⓐ. $\eta = \frac{Q_H}{Q_C}$
ⓑ. $\eta = \frac{W}{Q_H}$
ⓒ. $\eta = \frac{Q_H}{W}$
ⓓ. $\eta = \frac{Q_C}{Q_H}$
Correct Answer: $\eta = \frac{W}{Q_H}$
Explanation: Efficiency is the ratio of useful work done to heat absorbed from the hot reservoir. Since $W = Q_H – Q_C$, efficiency is also written as $\eta = 1 – \frac{Q_C}{Q_H}$.
277. A heat engine absorbs $3000 \, J$ of heat and rejects $2000 \, J$ to the sink. Find its efficiency.
ⓐ. 25%
ⓑ. 33%
ⓒ. 50%
ⓓ. 66%
Correct Answer: 33%
Explanation: Work done = $Q_H – Q_C = 3000 – 2000 = 1000 \, J$.$$\eta = \frac{W}{Q_H} = \frac{1000}{3000} = 0.333 = 33\%.$$
278. In an internal combustion engine, the hot source is:
ⓐ. The cooling water
ⓑ. The fuel-air mixture combustion
ⓒ. The piston
ⓓ. The exhaust gases
Correct Answer: The fuel-air mixture combustion
Explanation: In an internal combustion engine, the heat source is the burning of fuel-air mixture inside the cylinder. This provides energy to the working substance (gases), which expands and drives the piston.
279. Which statement is true for all heat engines?
ⓐ. They can achieve 100% efficiency
ⓑ. They always reject some heat to a cold reservoir
ⓒ. They operate without a working substance
ⓓ. They can work without a temperature difference
Correct Answer: They always reject some heat to a cold reservoir
Explanation: According to the Second Law of Thermodynamics, no heat engine can convert all input heat into work. Some heat must always be rejected to the sink.
280. The Otto cycle is the idealized cycle for:
ⓐ. Diesel engines
ⓑ. Steam turbines
ⓒ. Spark ignition engines
ⓓ. Refrigerators
Correct Answer: Spark ignition engines
Explanation: The Otto cycle models petrol engines (spark ignition). It consists of two isentropic processes and two isochoric processes. Diesel engines follow the Diesel cycle, steam turbines follow the Rankine cycle, and refrigerators are based on reverse Carnot or vapor-compression cycles.
281. The Carnot cycle is based on which principle?
ⓐ. Conservation of mass
ⓑ. Conservation of charge
ⓒ. Reversible processes between two heat reservoirs
ⓓ. Spontaneous heat transfer from cold to hot
Correct Answer: Reversible processes between two heat reservoirs
Explanation: The Carnot cycle is an ideal reversible cycle operating between two reservoirs at temperatures $T_H$ and $T_C$. It represents the maximum possible efficiency achievable by any heat engine working between those two temperatures.
282. Which of the following processes are part of the Carnot cycle?
ⓐ. Two isothermal and two adiabatic
ⓑ. Two isobaric and two isochoric
ⓒ. One isothermal, one adiabatic, one isobaric, one isochoric
ⓓ. Four adiabatic processes
Correct Answer: Two isothermal and two adiabatic
Explanation: The Carnot cycle consists of isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. This combination ensures a fully reversible cycle.
283. The efficiency of a Carnot engine operating between $T_H$ and $T_C$ is:
ⓐ. $\eta = \frac{Q_C}{Q_H}$
ⓑ. $\eta = \frac{W}{Q_H}$
ⓒ. $\eta = 1 – \frac{T_C}{T_H}$
ⓓ. $\eta = \frac{T_H}{T_C}$
Correct Answer: $\eta = 1 – \frac{T_C}{T_H}$
Explanation: Carnot efficiency depends only on the temperatures of the hot and cold reservoirs (absolute scale). It shows no real engine can exceed this efficiency.
284. A Carnot engine operates between a source at 600 K and sink at 300 K. Find its efficiency.
ⓐ. 25%
ⓑ. 40%
ⓒ. 50%
ⓓ. 60%
Correct Answer: 50%
Explanation: $$\eta = 1 – \frac{T_C}{T_H} = 1 – \frac{300}{600} = 1 – 0.5 = 0.5.$$
So, efficiency = 50%.
285. If the temperature of the sink in a Carnot engine is lowered, its efficiency will:
ⓐ. Decrease
ⓑ. Increase
ⓒ. Remain unchanged
ⓓ. Become zero
Correct Answer: Increase
Explanation: Efficiency of Carnot engine depends on $1 – \frac{T_C}{T_H}$. Reducing $T_C$ increases efficiency, as the difference between source and sink temperatures grows.
286. A Carnot engine takes in 1000 J of heat at 500 K and rejects heat at 300 K. Calculate the work output.
ⓐ. 200 J
ⓑ. 300 J
ⓒ. 400 J
ⓓ. 500 J
Correct Answer: 400 J
Explanation: Efficiency:$$\eta = 1 – \frac{300}{500} = 0.4.$$
$$W = \eta Q_H = 0.4 \times 1000 = 400 \, J.$$
287. The efficiency of a Carnot engine is always:
ⓐ. Greater than any real engine
ⓑ. Equal to all real engines
ⓒ. Less than all real engines
ⓓ. Independent of reservoir temperatures
Correct Answer: Greater than any real engine
Explanation: No real engine can surpass the efficiency of a Carnot engine because real processes involve irreversibilities like friction, turbulence, and non-ideal heat transfer.
288. A Carnot refrigerator working between 270 K and 300 K extracts 540 J of heat from the cold reservoir. Find the work input required.
ⓐ. 40 J
ⓑ. 50 J
ⓒ. 54 J
ⓓ. 60 J
Correct Answer: 60 J
Explanation: $$COP = \frac{T_C}{T_H – T_C} = \frac{270}{30} = 9.$$
$$COP = \frac{Q_C}{W} \Rightarrow W = \frac{Q_C}{COP} = \frac{540}{9} = 60 \, J.$$
289. The Carnot efficiency depends on:
ⓐ. Nature of working substance
ⓑ. Type of cycle used
ⓒ. Only on $T_H$ and $T_C$
ⓓ. Pressure of operation
Correct Answer: Only on $T_H$ and $T_C$
Explanation: Carnot efficiency is independent of the working substance or cycle details. It depends solely on the hot and cold reservoir temperatures, making it a universal standard.
290. A Carnot engine has efficiency 40% when operating between 500 K and $T_C$. Find the sink temperature.
ⓐ. 200 K
ⓑ. 250 K
ⓒ. 320 K
ⓓ. 300 K
Correct Answer: 300 K
Explanation: $$\eta = 1 – \frac{T_C}{T_H} \Rightarrow 0.4 = 1 – \frac{T_C}{500}.$$
$$\frac{T_C}{500} = 0.6 \Rightarrow T_C = 300 \, K.$$
291. Which thermodynamic cycle is the basis of most steam power plants?
ⓐ. Otto cycle
ⓑ. Diesel cycle
ⓒ. Rankine cycle
ⓓ. Carnot cycle
Correct Answer: Rankine cycle
Explanation: Modern steam power plants operate on the Rankine cycle, which uses water/steam as the working fluid. It consists of isobaric heating in a boiler, isentropic expansion in a turbine, isobaric heat rejection in a condenser, and isentropic compression in a pump.
292. The Otto cycle is the idealized cycle for:
ⓐ. Spark ignition (petrol) engines
ⓑ. Compression ignition (diesel) engines
ⓒ. Steam turbines
ⓓ. Gas turbines
Correct Answer: Spark ignition (petrol) engines
Explanation: The Otto cycle models petrol engines where ignition is caused by a spark. It consists of two adiabatic (isentropic) and two isochoric (constant volume) processes.
293. The Diesel cycle differs from the Otto cycle because:
ⓐ. Heat is supplied at constant pressure in the Diesel cycle
ⓑ. Heat is supplied at constant volume in the Diesel cycle
ⓒ. Work done is zero in the Diesel cycle
ⓓ. Diesel cycle does not involve adiabatic processes
Correct Answer: Heat is supplied at constant pressure in the Diesel cycle
Explanation: In the Otto cycle, heat addition occurs at constant volume, while in the Diesel cycle, heat addition occurs at constant pressure. This makes the Diesel engine more fuel-efficient.
294. The Brayton cycle is used in:
ⓐ. Jet engines and gas turbines
ⓑ. Steam power plants
ⓒ. Diesel trucks
ⓓ. Refrigerators
Correct Answer: Jet engines and gas turbines
Explanation: The Brayton cycle (or Joule cycle) is the ideal cycle for gas turbines and jet propulsion. It consists of two adiabatic and two isobaric processes.
295. A steam turbine in a thermal power plant produces 2000 kJ of work while receiving 5000 kJ of heat from the boiler. Calculate its efficiency.
ⓐ. 30%
ⓑ. 35%
ⓒ. 40%
ⓓ. 45%
Correct Answer: 40%
Explanation: $$\eta = \frac{W}{Q_H} = \frac{2000}{5000} = 0.4 = 40\%.$$
296. Why are Diesel engines generally more fuel-efficient than petrol engines?
ⓐ. Diesel has a lower calorific value
ⓑ. Diesel cycle has higher compression ratio
ⓒ. Otto cycle wastes less energy
ⓓ. Petrol burns slower
Correct Answer: Diesel cycle has higher compression ratio
Explanation: Diesel engines operate at higher compression ratios, leading to higher efficiency according to thermodynamic principles. Otto cycle engines (petrol) cannot reach such high compression due to knocking.
297. In a Rankine cycle, the component that converts liquid water into high-pressure steam is:
ⓐ. Condenser
ⓑ. Turbine
ⓒ. Boiler
ⓓ. Pump
Correct Answer: Boiler
Explanation: In the boiler, heat is supplied at nearly constant pressure to convert liquid water into high-pressure, high-temperature steam, which then expands in the turbine.
298. The efficiency of a Carnot engine between 800 K and 300 K is:
ⓐ. 50%
ⓑ. 55%
ⓒ. 60%
ⓓ. 62.5%
Correct Answer: 62.5%
Explanation: $$\eta = 1 – \frac{T_C}{T_H} = 1 – \frac{300}{800} = 1 – 0.375 = 0.625 = 62.5\%.$$
299. Which of the following engines is used in nuclear power plants for converting steam energy into electricity?
ⓐ. Diesel engine
ⓑ. Steam turbine
ⓒ. Jet engine
ⓓ. Gasoline engine
Correct Answer: Steam turbine
Explanation: Nuclear power plants generate heat through nuclear fission, producing steam. This steam drives turbines connected to electrical generators, following the Rankine cycle.
300. Which cycle is commonly used in aircraft gas turbines?
ⓐ. Otto cycle
ⓑ. Diesel cycle
ⓒ. Rankine cycle
ⓓ. Brayton cycle
Correct Answer: Brayton cycle
Explanation: Aircraft jet engines and modern gas turbines operate on the Brayton cycle, which involves isobaric heat addition, adiabatic expansion, isobaric heat rejection, and adiabatic compression.
You are now on Class 11 Physics MCQs – Chapter 12: Thermodynamics (Part 3). This section emphasizes the second law of thermodynamics, reversible and irreversible processes, heat engines, Carnot cycle, and efficiency of heat engines. These subtopics are highly significant for board exams and hold great weightage in JEE, NEET, and competitive tests. Across this chapter, we provide 500 MCQs with correct answers divided into 5 systematic parts. In Part 3, you will practice 100 more MCQs aimed at testing conceptual clarity as well as numerical aptitude.
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