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101. Boyle’s law states that for a fixed mass of gas at constant temperature:
ⓐ. $P \propto V$
ⓑ. $PV = \text{constant}$
ⓒ. $P \propto T$
ⓓ. $V \propto T$
Correct Answer: $PV = \text{constant}$
Explanation: Boyle’s law shows an inverse relationship between pressure and volume at constant temperature. If volume decreases, pressure increases proportionally, keeping their product constant. Other options represent Charles’s law or Gay-Lussac’s law.
102. If a gas occupies 4 L at 2 atm pressure, what volume will it occupy at 1 atm, assuming constant temperature?
ⓐ. 2 L
ⓑ. 4 L
ⓒ. 6 L
ⓓ. 8 L
Correct Answer: 8 L
Explanation: From Boyle’s law: $P_1V_1 = P_2V_2$. Substituting $2 \times 4 = 1 \times V_2$, we get $V_2 = 8 \, \text{L}$. Thus, halving pressure doubles the volume.
103. In a Boyle’s law experiment, a gas sample initially at 1 atm and 500 mL is compressed to 250 mL. What is the new pressure?
ⓐ. 0.5 atm
ⓑ. 1 atm
ⓒ. 2 atm
ⓓ. 4 atm
Correct Answer: 2 atm
Explanation: Using $P_1V_1 = P_2V_2$, $1 \times 500 = P_2 \times 250$. Solving gives $P_2 = 2 \, \text{atm}$. Pressure doubles when volume is halved at constant temperature.
104. Which type of graph is obtained when pressure is plotted against $1/V$ for a fixed mass of gas at constant temperature?
ⓐ. Straight line through origin
ⓑ. Parabola
ⓒ. Hyperbola
ⓓ. Circle
Correct Answer: Straight line through origin
Explanation: Since $P \propto 1/V$, plotting pressure against reciprocal of volume gives a straight line passing through the origin. Plotting $P$ against $V$ gives a rectangular hyperbola instead.
105. Which of the following best demonstrates Boyle’s law in real life?
ⓐ. Balloons expanding when heated
ⓑ. Syringe plunger moving when pressed
ⓒ. Increase in speed of gas molecules with temperature
ⓓ. Ice floating on water
Correct Answer: Syringe plunger moving when pressed
Explanation: When the plunger is pushed, the volume of air decreases, causing pressure to rise. This is a direct application of Boyle’s law. Heating a balloon involves Charles’s law, not Boyle’s.
106. A gas has a pressure of 5 atm when its volume is 10 L. What will be the pressure if the gas is compressed to 2 L at constant temperature?
ⓐ. 10 atm
ⓑ. 15 atm
ⓒ. 20 atm
ⓓ. 25 atm
Correct Answer: 25 atm
Explanation: Using $P_1V_1 = P_2V_2$, we have $5 \times 10 = P_2 \times 2$. So, $P_2 = 25 \, \text{atm}$. Pressure increases inversely with volume.
107. Which scientist first published the law of pressure-volume relationship (Boyle’s law)?
ⓐ. Robert Boyle
ⓑ. Jacques Charles
ⓒ. Amedeo Avogadro
ⓓ. Gay-Lussac
Correct Answer: Robert Boyle
Explanation: In 1662, Robert Boyle studied air under different pressures and volumes, formulating the law $PV = \text{constant}$. Charles studied volume-temperature relation, Avogadro volume-mole relation, and Gay-Lussac pressure-temperature relation.
108. If the volume of a gas increases three times at constant temperature, its pressure will:
ⓐ. Increase three times
ⓑ. Decrease three times
ⓒ. Increase nine times
ⓓ. Decrease nine times
Correct Answer: Decrease three times
Explanation: Since $P \propto 1/V$, if $V$ becomes 3 times larger, $P$ becomes 3 times smaller. For example, $P_1V_1 = P_2V_2$ gives $P_2 = P_1/3$.
109. In scuba diving, why is it dangerous to ascend quickly without exhaling?
ⓐ. Temperature decreases sharply.
ⓑ. Volume of air in lungs expands as pressure decreases, which may cause lung rupture.
ⓒ. Solubility of oxygen decreases.
ⓓ. The diver loses buoyancy.
Correct Answer: Volume of air in lungs expands as pressure decreases, which may cause lung rupture.
Explanation: As divers ascend, water pressure decreases. According to Boyle’s law, lung air expands. If the diver does not exhale, this expansion can damage lung tissues. This is a practical safety application of Boyle’s law.
110. If a sealed syringe containing 30 mL of air at 1 atm is compressed to 10 mL without changing temperature, what is the final pressure?
ⓐ. 1 atm
ⓑ. 2 atm
ⓒ. 3 atm
ⓓ. 5 atm
Correct Answer: 3 atm
Explanation: Using Boyle’s law, $P_1V_1 = P_2V_2$. Substituting $1 \times 30 = P_2 \times 10$, gives $P_2 = 3 \, \text{atm}$. Thus, reducing volume to one-third increases pressure three times.
111. Charles’s law states that for a fixed mass of gas at constant pressure:
ⓐ. $V \propto P$
ⓑ. $V \propto T$ (in Kelvin)
ⓒ. $P \propto \frac{1}{V}$
ⓓ. $V \propto \frac{1}{T}$
Correct Answer: $V \propto T$ (in Kelvin)
Explanation: Charles’s law says that the volume of a given mass of gas is directly proportional to its absolute temperature when pressure is constant. Thus, if temperature increases, volume also increases. The law only works with Kelvin, not Celsius.
112. Which of the following graphs correctly represents Charles’s law?
ⓐ. Pressure vs. Volume is a hyperbola
ⓑ. Volume vs. Temperature (K) is a straight line through origin
ⓒ. Pressure vs. Temperature is a straight line
ⓓ. Volume vs. Pressure is linear
Correct Answer: Volume vs. Temperature (K) is a straight line through origin
Explanation: Since $V \propto T$, plotting volume against absolute temperature gives a straight line through the origin. Other relationships belong to Boyle’s law or Gay-Lussac’s law.
113. A gas occupies 2 L at 300 K. If temperature is increased to 600 K at constant pressure, what is the new volume?
ⓐ. 1 L
ⓑ. 2 L
ⓒ. 3 L
ⓓ. 4 L
Correct Answer: 4 L
Explanation: Using $\frac{V_1}{T_1} = \frac{V_2}{T_2}$, we have $\frac{2}{300} = \frac{V_2}{600}$. Solving gives $V_2 = 4 \, \text{L}$. Volume doubles when absolute temperature doubles.
114. A balloon filled with air has a volume of 1.5 L at 27°C. What will be its volume at 127°C, if pressure remains constant?
ⓐ. 2.0 L
ⓑ. 2.5 L
ⓒ. 3.0 L
ⓓ. 1.0 L
Correct Answer: 2.0 L
Explanation: Convert to Kelvin: $T_1 = 27 + 273 = 300 \, K$, $T_2 = 127 + 273 = 400 \, K$. Using Charles’s law: $\frac{1.5}{300} = \frac{V_2}{400}$. Solving gives $V_2 = 2.0 \, \text{L}$.
115. Which scientist discovered the relationship now known as Charles’s law?
ⓐ. Robert Boyle
ⓑ. Jacques Charles
ⓒ. Amedeo Avogadro
ⓓ. Gay-Lussac
Correct Answer: Jacques Charles
Explanation: In 1787, Jacques Charles studied gases and found that volume increases proportionally with temperature at constant pressure. His unpublished work was later confirmed by Gay-Lussac.
116. According to Charles’s law, if absolute temperature of a gas decreases, its volume:
ⓐ. Increases
ⓑ. Decreases
ⓒ. Remains constant
ⓓ. Becomes zero
Correct Answer: Decreases
Explanation: Charles’s law states $V \propto T$. If $T$ decreases, $V$ also decreases. Extrapolating this linearly leads to zero volume at absolute zero (0 K), a concept later refined by kinetic theory.
117. A gas has a volume of 500 mL at 20°C. What will be its volume at 0°C at constant pressure?
ⓐ. 333 mL
ⓑ. 450 mL
ⓒ. 400 mL
ⓓ. 273 mL
Correct Answer: 450 mL
Explanation: Convert to Kelvin: $T_1 = 293 K, T_2 = 273 K$. Using $\frac{V_1}{T_1} = \frac{V_2}{T_2}$, $\frac{500}{293} = \frac{V_2}{273}$. Solving gives $V_2 \approx 466 \, \text{mL}$. Closest option is 450 mL.
118. Why must Charles’s law be expressed in Kelvin instead of Celsius?
ⓐ. Celsius is not scientific.
ⓑ. Celsius can give negative temperatures, leading to negative volumes.
ⓒ. Kelvin scale is larger than Celsius.
ⓓ. Kelvin eliminates molecular mass.
Correct Answer: Celsius can give negative temperatures, leading to negative volumes.
Explanation: Volume cannot be negative, but Celsius scale allows negative values. The Kelvin scale starts from absolute zero (0 K), where molecular motion theoretically ceases. This makes proportionality accurate.
119. A balloon at constant pressure shrinks in volume when cooled. Which molecular explanation fits Charles’s law?
ⓐ. Molecules move faster at low temperature.
ⓑ. Molecules lose kinetic energy, reducing frequency and force of collisions.
ⓒ. Molecules gain mass when cooled.
ⓓ. Number of molecules decreases.
Correct Answer: Molecules lose kinetic energy, reducing frequency and force of collisions.
Explanation: As temperature decreases, average molecular kinetic energy decreases. Fewer collisions with container walls at lower force cause volume to shrink at constant pressure, consistent with Charles’s law.
120. A sample of gas has a volume of 250 mL at 300 K. To maintain constant pressure, what temperature is needed for the volume to expand to 500 mL?
ⓐ. 450 K
ⓑ. 550 K
ⓒ. 600 K
ⓓ. 700 K
Correct Answer: 600 K
Explanation: Using $\frac{V_1}{T_1} = \frac{V_2}{T_2}$, we have $\frac{250}{300} = \frac{500}{T_2}$. Solving gives $T_2 = 600 \, K$. Thus, doubling volume requires doubling absolute temperature.
121. Avogadro’s law states that:
ⓐ. Equal volumes of gases at same temperature and pressure contain equal number of molecules.
ⓑ. Equal masses of gases at same pressure contain equal number of molecules.
ⓒ. Volume is directly proportional to temperature.
ⓓ. Pressure is inversely proportional to volume.
Correct Answer: Equal volumes of gases at same temperature and pressure contain equal number of molecules.
Explanation: Avogadro’s law explains the relation between volume and number of molecules at constant temperature and pressure. This principle allowed chemists to define the mole concept and calculate molar volumes.
122. The molar volume of an ideal gas at STP (Standard Temperature and Pressure, 273 K and 1 atm) is:
ⓐ. 11.2 L
ⓑ. 22.4 L
ⓒ. 44.8 L
ⓓ. 1.0 L
Correct Answer: 22.4 L
Explanation: According to Avogadro’s law, one mole of any ideal gas occupies 22.4 L at STP. This volume is independent of gas type, as long as the conditions are the same.
123. If 2 L of hydrogen and 2 L of oxygen are measured at the same temperature and pressure, then:
ⓐ. Both have different number of molecules.
ⓑ. Both contain equal number of molecules.
ⓒ. Oxygen has double the number of molecules.
ⓓ. Hydrogen has fewer molecules.
Correct Answer: Both contain equal number of molecules.
Explanation: Avogadro’s law states that equal volumes of gases at the same conditions have equal numbers of molecules, regardless of chemical identity. Hence, both gases contain equal molecules.
124. Which of the following chemical reactions demonstrates Avogadro’s law?
ⓐ. $2H_2 + O_2 \rightarrow 2H_2O$
ⓑ. $H_2 + Cl_2 \rightarrow 2HCl$
ⓒ. $N_2 + 3H_2 \rightarrow 2NH_3$
ⓓ. All of the above
Correct Answer: All of the above
Explanation: In each case, gas volumes can be related stoichiometrically. Avogadro’s law allows chemists to interpret chemical equations in terms of both number of molecules and volumes of gases at STP.
125. If 3 moles of a gas occupy 67.2 L at STP, then the volume of 1 mole of the gas is:
ⓐ. 11.2 L
ⓑ. 22.4 L
ⓒ. 44.8 L
ⓓ. 33.6 L
Correct Answer: 22.4 L
Explanation: Dividing total volume by number of moles, $V_m = \frac{67.2}{3} = 22.4 \, L$. This confirms Avogadro’s principle of molar volume at STP.
126. Which scientist is credited with formulating Avogadro’s law?
ⓐ. Robert Boyle
ⓑ. Jacques Charles
ⓒ. Amedeo Avogadro
ⓓ. Gay-Lussac
Correct Answer: Amedeo Avogadro
Explanation: In 1811, Amedeo Avogadro proposed that equal volumes of gases at same temperature and pressure contain equal number of molecules. His hypothesis clarified gas laws and molecular masses.
127. The density of a gas at STP can be calculated using Avogadro’s law and molar mass. Which is the formula?
ⓐ. $d = \frac{M}{22.4}$ g/L
ⓑ. $d = \frac{22.4}{M}$ g/L
ⓒ. $d = \frac{M}{V}$ g/L
ⓓ. $d = M \times 22.4$ g/L
Correct Answer: $d = \frac{M}{22.4}$ g/L
Explanation: One mole of gas occupies 22.4 L at STP. Density is mass/volume, so density = molar mass (M) / molar volume (22.4 L). This relation is widely used in chemical calculations.
128. If a gas sample contains 6.022 × $10^{23}$ molecules at STP, its volume is:
ⓐ. 1 L
ⓑ. 11.2 L
ⓒ. 22.4 L
ⓓ. 44.8 L
Correct Answer: 22.4 L
Explanation: By definition, one mole of any ideal gas contains Avogadro’s number of molecules and occupies 22.4 L at STP. Thus, a sample with 6.022 × $10^{23}$ molecules has volume 22.4 L.
129. Avogadro’s law provides direct evidence for which concept?
ⓐ. Atomic number
ⓑ. Conservation of energy
ⓒ. Atomic hypothesis and existence of molecules
ⓓ. Wave-particle duality
Correct Answer: Atomic hypothesis and existence of molecules
Explanation: Avogadro’s law strengthened the idea that matter consists of discrete particles. By showing equal volumes contain equal numbers of molecules, it provided evidence for the molecular basis of matter.
130. Which experimental observation best supports Avogadro’s law?
ⓐ. Equal volumes of different gases have different densities.
ⓑ. Equal volumes of different gases at same conditions give simple ratios in reactions.
ⓒ. Volume is inversely proportional to pressure.
ⓓ. Volume is directly proportional to Kelvin temperature.
Correct Answer: Equal volumes of different gases at same conditions give simple ratios in reactions.
Explanation: For example, 2 L of hydrogen reacting with 1 L of oxygen to form 2 L of steam shows simple whole-number volume ratios, explained by Avogadro’s law that equal volumes contain equal numbers of molecules.
131. Which of the following is a basic assumption of the kinetic theory of gases?
ⓐ. Gas molecules exert strong attractive forces on each other.
ⓑ. Gas molecules are in constant random motion.
ⓒ. Gas molecules move in fixed circular paths.
ⓓ. Gas molecules lose kinetic energy in collisions.
Correct Answer: Gas molecules are in constant random motion.
Explanation: One assumption of kinetic theory is that molecules move continuously and randomly in all directions. Their collisions are elastic, so kinetic energy is conserved. Strong attractions (A) and inelastic collisions (D) are not assumed in the ideal gas model.
132. According to kinetic theory, the volume of individual gas molecules compared to the total volume of the gas is:
ⓐ. Very large
ⓑ. Negligible
ⓒ. Equal
ⓓ. Infinite
Correct Answer: Negligible
Explanation: Gas molecules are treated as point masses with negligible size compared to the container volume. This assumption simplifies calculations and is valid at low pressure. At high pressure, molecular volume cannot be ignored, leading to deviations.
133. Which assumption of kinetic theory explains why pressure of a gas arises?
ⓐ. Molecules exert no forces on each other.
ⓑ. Gas molecules collide elastically with container walls.
ⓒ. Molecules move in circular orbits.
ⓓ. Gas molecules have variable mass.
Correct Answer: Gas molecules collide elastically with container walls.
Explanation: Each collision with a wall changes molecular momentum, exerting force. The total force per unit area of collisions results in pressure. Elastic collisions conserve kinetic energy, maintaining pressure over time.
134. Kinetic theory assumes that collisions between gas molecules are:
ⓐ. Perfectly inelastic
ⓑ. Perfectly elastic
ⓒ. Explosive
ⓓ. Partially elastic
Correct Answer: Perfectly elastic
Explanation: Elastic collisions ensure no net loss of kinetic energy, though energy may be exchanged between molecules. This assumption aligns with the First Law of Thermodynamics (energy conservation).
135. Which assumption connects kinetic theory to absolute temperature?
ⓐ. Average kinetic energy of molecules is proportional to Kelvin temperature.
ⓑ. Molecular velocity is constant.
ⓒ. Gas molecules exert strong intermolecular forces.
ⓓ. Molecular energy is independent of temperature.
Correct Answer: Average kinetic energy of molecules is proportional to Kelvin temperature.
Explanation: Kinetic theory states $\langle E_k \rangle = \tfrac{3}{2}k_B T$. Thus, temperature is a direct measure of molecular kinetic energy, bridging microscopic motion and macroscopic thermodynamics.
136. Which of the following is NOT an assumption of kinetic theory?
ⓐ. Gas molecules are point particles.
ⓑ. Intermolecular forces are negligible.
ⓒ. Gas molecules move in straight lines until collisions occur.
ⓓ. Gas molecules stick together after collisions.
Correct Answer: Gas molecules stick together after collisions.
Explanation: Kinetic theory assumes no sticking; collisions are perfectly elastic. Sticking would reduce kinetic energy and contradict the theory. Straight-line motion (C) occurs between collisions, as assumed.
137. In kinetic theory, what is assumed about the time of collision between two molecules?
ⓐ. Very large compared to time between collisions
ⓑ. Negligible compared to time between collisions
ⓒ. Equal to time between collisions
ⓓ. Infinite
Correct Answer: Negligible compared to time between collisions
Explanation: The duration of a collision is considered extremely small compared to the free time molecules travel. This simplifies the analysis of gas behavior by treating collisions as instantaneous.
138. According to kinetic theory, the distribution of molecular speeds in a gas:
ⓐ. Is the same for all gases at the same temperature.
ⓑ. Depends only on pressure.
ⓒ. Is independent of temperature.
ⓓ. Is constant for all masses of molecules.
Correct Answer: Is the same for all gases at the same temperature.
Explanation: While individual molecular speeds differ due to mass, the distribution shape is governed by temperature. Lighter molecules move faster, but the Maxwell–Boltzmann distribution applies to all gases at a given $T$.
139. Which assumption of kinetic theory is violated when gases show liquefaction at high pressure?
ⓐ. Gas molecules are in random motion.
ⓑ. Collisions are elastic.
ⓒ. Intermolecular forces are negligible.
ⓓ. Average kinetic energy is proportional to temperature.
Correct Answer: Intermolecular forces are negligible.
Explanation: Real gases deviate because at high pressure and low temperature, attractive forces become significant, leading to condensation. This violates the ideal assumption of negligible intermolecular forces.
140. The assumption that gas molecules move in straight lines between collisions implies:
ⓐ. Molecules have no acceleration except during collisions.
ⓑ. Molecules continuously slow down.
ⓒ. Molecules orbit around the nucleus.
ⓓ. Molecules are bound in fixed positions.
Correct Answer: Molecules have no acceleration except during collisions.
Explanation: Kinetic theory assumes no external forces act except during collisions. Therefore, molecules move with constant velocity in straight lines until they collide with other molecules or container walls.
141. According to kinetic theory, the pressure exerted by a gas in a container is due to:
ⓐ. Gravitational pull on the molecules
ⓑ. Elastic collisions of molecules with container walls
ⓒ. Repulsion between molecules
ⓓ. Constant acceleration of molecules
Correct Answer: Elastic collisions of molecules with container walls
Explanation: Gas molecules move randomly and collide elastically with the container walls. Each collision transfers momentum, and the cumulative effect of these collisions per unit area produces measurable pressure.
142. Which formula relates pressure $P$, number density $n$, molecular mass $m$, and mean square speed $\langle v^2 \rangle$?
ⓐ. $P = \frac{1}{2}nm\langle v^2 \rangle$
ⓑ. $P = \frac{1}{3}nm\langle v^2 \rangle$
ⓒ. $P = nm\langle v^2 \rangle$
ⓓ. $P = \frac{2}{3}nm\langle v^2 \rangle$
Correct Answer: $P = \frac{1}{3}nm\langle v^2 \rangle$
Explanation: From kinetic theory, $P = \frac{1}{3} \rho \langle v^2 \rangle$ where $\rho = nm$ is mass density of the gas. The factor $\frac{1}{3}$ arises because motion occurs in three dimensions, distributing momentum equally.
143. The relation between average kinetic energy per molecule and temperature is:
ⓐ. $\langle E_k \rangle = k_B T$
ⓑ. $\langle E_k \rangle = \frac{1}{2}k_B T$
ⓒ. $\langle E_k \rangle = \frac{3}{2}k_B T$
ⓓ. $\langle E_k \rangle = \frac{5}{2}k_B T$
Correct Answer: $\langle E_k \rangle = \frac{3}{2}k_B T$
Explanation: Kinetic theory proves that average translational kinetic energy of a molecule depends only on absolute temperature: $\langle E_k \rangle = \frac{3}{2}k_B T$. This links microscopic motion to macroscopic temperature.
144. Which expression relates the rms speed of molecules to pressure and density?
ⓐ. $v_{rms} = \sqrt{\frac{P}{\rho}}$
ⓑ. $v_{rms} = \sqrt{\frac{2P}{\rho}}$
ⓒ. $v_{rms} = \sqrt{\frac{3P}{\rho}}$
ⓓ. $v_{rms} = \sqrt{\frac{P}{3\rho}}$
Correct Answer: $v_{rms} = \sqrt{\frac{3P}{\rho}}$
Explanation: Using $P = \tfrac{1}{3}\rho \langle v^2 \rangle$, we get $\langle v^2 \rangle = \frac{3P}{\rho}$. Taking the square root gives rms speed: $v_{rms} = \sqrt{\frac{3P}{\rho}}$.
145. In the derivation of gas pressure, why is the factor $\frac{1}{3}$ included?
ⓐ. Only one-third of molecules move in the x-direction at a time.
ⓑ. Molecules move only in one plane.
ⓒ. To account for intermolecular forces.
ⓓ. Because pressure is isotropic in liquids, not gases.
Correct Answer: Only one-third of molecules move in the x-direction at a time.
Explanation: Motion is random and equally distributed among the x, y, and z directions. Thus, the mean square velocity along one axis is one-third of the total mean square velocity, leading to the $\frac{1}{3}$ factor.
146. The ideal gas equation $PV = Nk_B T$ is consistent with kinetic theory because:
ⓐ. It assumes molecules are stationary.
ⓑ. It relates pressure to average molecular kinetic energy.
ⓒ. It neglects temperature.
ⓓ. It uses empirical values only.
Correct Answer: It relates pressure to average molecular kinetic energy.
Explanation: From kinetic theory, $P = \tfrac{1}{3} nm \langle v^2 \rangle$. Rearranging gives $PV = Nk_B T$, connecting macroscopic properties (P, V, T) with microscopic kinetic energy.
147. For 1 mole of an ideal gas, the total translational kinetic energy is:
ⓐ. $\frac{1}{2}RT$
ⓑ. $RT$
ⓒ. $\frac{3}{2}RT$
ⓓ. $3RT$
Correct Answer: $\frac{3}{2}RT$
Explanation: For each molecule, $\langle E_k \rangle = \tfrac{3}{2}k_B T$. Multiplying by Avogadro’s number $N_A$ gives $U = \tfrac{3}{2}RT$. Thus, one mole of an ideal gas has kinetic energy proportional to temperature.
148. Which molecular property is directly measured by temperature according to kinetic theory?
ⓐ. Momentum
ⓑ. Velocity
ⓒ. Average kinetic energy
ⓓ. Force
Correct Answer: Average kinetic energy
Explanation: Temperature is a measure of average translational kinetic energy of molecules. Higher temperature means molecules move faster, while pressure and volume are influenced by molecular collisions.
149. What happens to the pressure of a gas if the rms speed of its molecules doubles at constant volume?
ⓐ. Pressure remains the same
ⓑ. Pressure doubles
ⓒ. Pressure quadruples
ⓓ. Pressure halves
Correct Answer: Pressure quadruples
Explanation: Pressure is proportional to mean square speed: $P \propto \langle v^2 \rangle$. If rms speed doubles, $v_{rms}^2$ increases by 4, so pressure increases four times at constant volume.
150. In kinetic theory derivation, the temperature of a gas is defined in terms of:
ⓐ. Average momentum per molecule
ⓑ. Average kinetic energy per molecule
ⓒ. Total number of collisions per second
ⓓ. Volume occupied by molecules
Correct Answer: Average kinetic energy per molecule
Explanation: The absolute temperature $T$ is defined via $\langle E_k \rangle = \frac{3}{2}k_B T$. Thus, temperature is a macroscopic measure of the average microscopic energy of gas molecules.
151. The Maxwell–Boltzmann distribution describes:
ⓐ. Distribution of molecular masses in a gas
ⓑ. Distribution of molecular speeds at a given temperature
ⓒ. Distribution of gas pressures in different containers
ⓓ. Distribution of electron orbits in an atom
Correct Answer: Distribution of molecular speeds at a given temperature
Explanation: Maxwell–Boltzmann distribution gives the probability of molecules having certain speeds in a gas. It shows that not all molecules move at the same speed, but speeds are spread around a most probable value depending on temperature.
152. In the Maxwell–Boltzmann speed distribution curve, the area under the curve represents:
ⓐ. Average molecular speed
ⓑ. Number of molecules per unit volume
ⓒ. Total number of molecules in the sample
ⓓ. Pressure of the gas
Correct Answer: Total number of molecules in the sample
Explanation: The distribution curve plots number of molecules vs. molecular speed. The integral (area under the curve) equals the total number of molecules, just like a probability distribution in statistics.
153. Which of the following speeds is NOT defined in the context of Maxwell–Boltzmann distribution?
ⓐ. Most probable speed
ⓑ. Average speed
ⓒ. Root mean square (rms) speed
ⓓ. Escape speed
Correct Answer: Escape speed
Explanation: Maxwell–Boltzmann distribution defines three characteristic speeds—most probable ($v_{mp}$), average ($v_{avg}$), and rms ($v_{rms}$). Escape speed refers to celestial bodies, not molecular speed distribution.
154. The most probable speed of gas molecules is given by:
ⓐ. $v_{mp} = \sqrt{\frac{2RT}{M}}$
ⓑ. $v_{mp} = \sqrt{\frac{3RT}{M}}$
ⓒ. $v_{mp} = \sqrt{\frac{RT}{M}}$
ⓓ. $v_{mp} = \sqrt{\frac{8RT}{\pi M}}$
Correct Answer: $v_{mp} = \sqrt{\frac{2RT}{M}}$
Explanation: From Maxwell–Boltzmann distribution, the speed at which the distribution curve peaks (most probable speed) is $v_{mp} = \sqrt{\frac{2RT}{M}}$. The rms and average speeds have different formulae.
155. The average speed of molecules in Maxwell–Boltzmann distribution is:
ⓐ. $v_{avg} = \sqrt{\frac{2RT}{M}}$
ⓑ. $v_{avg} = \sqrt{\frac{3RT}{M}}$
ⓒ. $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$
ⓓ. $v_{avg} = \sqrt{\frac{RT}{M}}$
Correct Answer: $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$
Explanation: The average molecular speed is calculated using integration of the distribution function, resulting in $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$. This lies between the most probable and rms speeds.
156. Which of the following is the correct order of characteristic molecular speeds?
ⓐ. $v_{rms} > v_{avg} > v_{mp}$
ⓑ. $v_{rms} > v_{mp} > v_{avg}$
ⓒ. $v_{avg} > v_{rms} > v_{mp}$
ⓓ. $v_{mp} > v_{avg} > v_{rms}$
Correct Answer: $v_{rms} > v_{avg} > v_{mp}$
Explanation: For gases, $v_{mp} = \sqrt{\frac{2RT}{M}}$, $v_{avg} = \sqrt{\frac{8RT}{\pi M}}$, and $v_{rms} = \sqrt{\frac{3RT}{M}}$. Numerically, $v_{rms} > v_{avg} > v_{mp}$.
157. What happens to the Maxwell–Boltzmann distribution curve when temperature increases?
ⓐ. The peak shifts to lower speed and narrows.
ⓑ. The peak shifts to higher speed and flattens.
ⓒ. The peak remains fixed but area decreases.
ⓓ. The curve disappears completely.
Correct Answer: The peak shifts to higher speed and flattens.
Explanation: At higher temperature, molecules have more kinetic energy, so the distribution broadens and most probable speed increases. The curve flattens because molecules are spread over a wider speed range.
158. For a fixed gas, what happens to the Maxwell–Boltzmann distribution when molar mass increases at constant temperature?
ⓐ. Distribution shifts toward lower speeds.
ⓑ. Distribution shifts toward higher speeds.
ⓒ. Distribution remains unchanged.
ⓓ. Distribution area decreases.
Correct Answer: Distribution shifts toward lower speeds.
Explanation: Heavier molecules move more slowly at the same temperature, since $v \propto \frac{1}{\sqrt{M}}$. Thus, the curve shifts left as molar mass increases.
159. The Maxwell–Boltzmann distribution provides a molecular basis for which thermodynamic law?
ⓐ. Zeroth Law
ⓑ. First Law
ⓒ. Second Law
ⓓ. Third Law
Correct Answer: Second Law
Explanation: The Second Law of Thermodynamics, concerning entropy and statistical probability, is explained by Maxwell–Boltzmann distribution. Systems evolve toward states with the most probable molecular distributions, i.e., maximum entropy.
160. Which experimental observation validates the Maxwell–Boltzmann distribution of molecular speeds?
ⓐ. Brownian motion
ⓑ. Diffusion rates of gases (Graham’s law)
ⓒ. Variation of pressure with volume (Boyle’s law)
ⓓ. Phase transitions of solids
Correct Answer: Diffusion rates of gases (Graham’s law)
Explanation: Graham’s law of diffusion shows that lighter gases diffuse faster than heavier ones, consistent with Maxwell–Boltzmann’s prediction that average molecular speed is inversely proportional to the square root of molar mass.
161. The law of equipartition of energy states that:
ⓐ. Energy is always conserved in chemical reactions.
ⓑ. Each quadratic degree of freedom contributes $\tfrac{1}{2}k_B T$ per molecule to energy.
ⓒ. Every particle has the same total energy at all times.
ⓓ. Energy distribution is unequal among different modes.
Correct Answer: Each quadratic degree of freedom contributes $\tfrac{1}{2}k_B T$ per molecule to energy.
Explanation: Equipartition theorem states that for each independent quadratic term in energy (like translational, rotational, vibrational motion), the average contribution is $\tfrac{1}{2}k_B T$ per molecule. This allows prediction of heat capacities of gases.
162. For a monatomic ideal gas, the total average energy per molecule according to equipartition theorem is:
ⓐ. $\tfrac{1}{2}k_B T$
ⓑ. $k_B T$
ⓒ. $\tfrac{3}{2}k_B T$
ⓓ. $3k_B T$
Correct Answer: $\tfrac{3}{2}k_B T$
Explanation: A monatomic gas has only 3 translational degrees of freedom (x, y, z motion). Thus, energy per molecule = $3 \times \tfrac{1}{2}k_B T = \tfrac{3}{2}k_B T$.
163. For a diatomic molecule (ignoring vibrational modes), how many degrees of freedom are considered in equipartition theorem?
ⓐ. 2
ⓑ. 3
ⓒ. 5
ⓓ. 6
Correct Answer: 5
Explanation: A diatomic molecule has 3 translational and 2 rotational degrees of freedom (rotation about two axes perpendicular to the bond). Hence, total = 5 degrees. Vibrations add more at higher temperatures.
164. What is the molar heat capacity at constant volume ($C_v$) of a monatomic ideal gas according to equipartition theorem?
ⓐ. $\tfrac{1}{2}R$
ⓑ. $\tfrac{3}{2}R$
ⓒ. $\tfrac{5}{2}R$
ⓓ. $3R$
Correct Answer: $\tfrac{3}{2}R$
Explanation: Internal energy per mole = $\tfrac{3}{2}RT$. Differentiating with respect to temperature gives $C_v = \tfrac{3}{2}R$. This matches experimental results for monatomic gases like helium and argon.
165. According to equipartition theorem, what is the molar heat capacity at constant pressure ($C_p$) for a monatomic ideal gas?
ⓐ. $\tfrac{3}{2}R$
ⓑ. $\tfrac{5}{2}R$
ⓒ. $\tfrac{7}{2}R$
ⓓ. $3R$
Correct Answer: $\tfrac{5}{2}R$
Explanation: For ideal gases, $C_p = C_v + R$. Since $C_v = \tfrac{3}{2}R$, we get $C_p = \tfrac{5}{2}R$. The ratio $\gamma = C_p/C_v = 5/3$ for monatomic gases.
166. For a diatomic gas at room temperature, neglecting vibrations, the molar $C_v$ predicted by equipartition theorem is:
ⓐ. $\tfrac{3}{2}R$
ⓑ. $\tfrac{5}{2}R$
ⓒ. $\tfrac{7}{2}R$
ⓓ. $3R$
Correct Answer: $\tfrac{5}{2}R$
Explanation: A diatomic molecule has 5 degrees of freedom (3 translational + 2 rotational). Internal energy per mole = $\tfrac{5}{2}RT$. Differentiating gives $C_v = \tfrac{5}{2}R$.
167. The failure of equipartition theorem at very low temperatures is explained by:
ⓐ. Newton’s laws
ⓑ. Quantum theory of energy levels
ⓒ. Boyle’s law
ⓓ. Avogadro’s law
Correct Answer: Quantum theory of energy levels
Explanation: Equipartition assumes continuous energy states, but quantum mechanics shows energy levels are discrete. At low temperatures, higher modes (like vibration, rotation) cannot be excited, so heat capacities are smaller than classical predictions.
168. The ratio of specific heats ($\gamma = C_p/C_v$) for a diatomic gas at moderate temperature is approximately:
ⓐ. 1.33
ⓑ. 1.40
ⓒ. 1.67
ⓓ. 2.00
Correct Answer: 1.40
Explanation: For diatomic gases (ignoring vibrations), $C_v = \tfrac{5}{2}R$, $C_p = \tfrac{7}{2}R$. Thus, $\gamma = \frac{C_p}{C_v} = \frac{7/2}{5/2} = 1.40$. This matches experimental data for O₂ and N₂ at room temperature.
169. Equipartition theorem helps explain which property of gases?
ⓐ. Pressure
ⓑ. Brownian motion
ⓒ. Specific heat capacity
ⓓ. Diffusion
Correct Answer: Specific heat capacity
Explanation: The theorem relates energy contributions from degrees of freedom to temperature, predicting specific heat capacities. Monatomic gases have $C_v = \tfrac{3}{2}R$, diatomic gases $\tfrac{5}{2}R$, etc. This links microscopic motion to macroscopic thermodynamic property.
170. Which practical application directly uses the law of equipartition of energy?
ⓐ. Predicting conductivity of metals
ⓑ. Calculating molar heat capacities of gases
ⓒ. Explaining buoyancy of liquids
ⓓ. Understanding nuclear binding energy
Correct Answer: Calculating molar heat capacities of gases
Explanation: The law of equipartition allows us to compute $C_v$ and $C_p$ for gases by counting degrees of freedom. This is fundamental in thermodynamics and explains why heat capacities differ for monatomic, diatomic, and polyatomic gases.
171. For a monatomic ideal gas, which degrees of freedom contribute to its energy?
ⓐ. 1 translational
ⓑ. 2 rotational
ⓒ. 3 translational
ⓓ. 3 translational + 3 rotational
Correct Answer: 3 translational
Explanation: Monatomic gases (like He, Ar, Ne) can only move in three perpendicular directions (x, y, z). Hence, they have 3 translational degrees of freedom. Rotations are not considered because the molecule is spherical and rotation does not add kinetic energy in this case.
172. According to the law of equipartition, the average energy per molecule of a monatomic ideal gas is:
ⓐ. $\tfrac{1}{2}k_B T$
ⓑ. $k_B T$
ⓒ. $\tfrac{3}{2}k_B T$
ⓓ. $\tfrac{5}{2}k_B T$
Correct Answer: $\tfrac{3}{2}k_B T$
Explanation: Each quadratic degree of freedom contributes $\tfrac{1}{2}k_B T$. A monatomic gas has 3 translational degrees, so energy per molecule is $3 \times \tfrac{1}{2}k_B T = \tfrac{3}{2}k_B T$.
173. For 1 mole of a monatomic gas at temperature $T$, the total internal energy is:
ⓐ. $U = \tfrac{3}{2}RT$
ⓑ. $U = \tfrac{5}{2}RT$
ⓒ. $U = 3RT$
ⓓ. $U = RT$
Correct Answer: $U = \tfrac{3}{2}RT$
Explanation: Multiplying the per-molecule energy by Avogadro’s number gives $U = N_A \times \tfrac{3}{2}k_B T = \tfrac{3}{2}RT$. This matches thermodynamic calculations of monatomic gas internal energy.
174. A diatomic gas molecule (ignoring vibrations) has how many degrees of freedom?
ⓐ. 3
ⓑ. 5
ⓒ. 6
ⓓ. 7
Correct Answer: 5
Explanation: A diatomic molecule has 3 translational (x, y, z) and 2 rotational (about axes perpendicular to the bond). Total = 5 degrees of freedom. Vibrational modes activate only at high temperatures.
175. The average energy per molecule of a diatomic gas at moderate temperatures (ignoring vibrations) is:
ⓐ. $\tfrac{3}{2}k_B T$
ⓑ. $\tfrac{5}{2}k_B T$
ⓒ. $2k_B T$
ⓓ. $3k_B T$
Correct Answer: $\tfrac{5}{2}k_B T$
Explanation: With 5 degrees of freedom, energy per molecule = $5 \times \tfrac{1}{2}k_B T = \tfrac{5}{2}k_B T$. This accounts for translational and rotational motions.
176. For 1 mole of a diatomic gas (ignoring vibrations), the internal energy is:
ⓐ. $U = \tfrac{3}{2}RT$
ⓑ. $U = \tfrac{5}{2}RT$
ⓒ. $U = 3RT$
ⓓ. $U = \tfrac{7}{2}RT$
Correct Answer: $U = \tfrac{5}{2}RT$
Explanation: Each degree of freedom contributes $\tfrac{1}{2}RT$ per mole. For 5 degrees of freedom, total energy = $\tfrac{5}{2}RT$. This explains the higher heat capacity of diatomic gases compared to monatomic gases.
177. A triatomic linear molecule like $CO_2$ has how many degrees of freedom (ignoring vibrations)?
ⓐ. 5
ⓑ. 6
ⓒ. 7
ⓓ. 8
Correct Answer: 6
Explanation: Linear molecules have 3 translational + 2 rotational + 1 rotational along the bond axis negligible (ignored). So effectively, 5 degrees at low T, but full consideration gives 6 (with some vibrational modes). At higher T, vibrations activate adding more degrees.
178. For a linear triatomic molecule at high temperature where vibrational modes are active, the average energy per molecule becomes:
ⓐ. $\tfrac{5}{2}k_B T$
ⓑ. $3k_B T$
ⓒ. $\tfrac{7}{2}k_B T$
ⓓ. $4k_B T$
Correct Answer: $\tfrac{7}{2}k_B T$
Explanation: A linear triatomic molecule has 3 translational + 2 rotational + 2 vibrational quadratic terms (each vibration has 2 DOF: potential + kinetic). Total = 7 degrees, giving average energy = $\tfrac{7}{2}k_B T$.
179. The molar heat capacity at constant volume, $C_v$, for a diatomic gas (ignoring vibrations) is:
ⓐ. $\tfrac{3}{2}R$
ⓑ. $\tfrac{5}{2}R$
ⓒ. $\tfrac{7}{2}R$
ⓓ. $3R$
Correct Answer: $\tfrac{5}{2}R$
Explanation: Internal energy per mole is $\tfrac{5}{2}RT$. Differentiating with respect to $T$, $C_v = \tfrac{5}{2}R$. This is larger than the monatomic value due to extra rotational degrees.
180. Calculate the internal energy of 2 moles of a monatomic ideal gas at 300 K.
ⓐ. $U = 900R$
ⓑ. $U = 600R$
ⓒ. $U = 300R$
ⓓ. $U = 150R$
Correct Answer: $U = 600R$
Explanation: For monatomic gas, $U = \tfrac{3}{2}nRT$. Substituting $n=2, T=300K$: $U = \tfrac{3}{2} \times 2 \times 300R = 600R$. This formula applies directly from equipartition and degrees of freedom.
181. For a monatomic ideal gas, the molar heat capacity at constant volume ($C_v$) is:
ⓐ. $R$
ⓑ. $\tfrac{3}{2}R$
ⓒ. $\tfrac{5}{2}R$
ⓓ. $3R$
Correct Answer: $\tfrac{3}{2}R$
Explanation: From equipartition, a monatomic gas has 3 translational degrees of freedom. Each contributes $\tfrac{1}{2}RT$ per mole, so total internal energy $U = \tfrac{3}{2}RT$. Differentiating gives $C_v = \tfrac{3}{2}R$.
182. For a diatomic ideal gas (neglecting vibration), the molar heat capacity at constant pressure ($C_p$) is:
ⓐ. $\tfrac{3}{2}R$
ⓑ. $\tfrac{5}{2}R$
ⓒ. $\tfrac{7}{2}R$
ⓓ. $\tfrac{9}{2}R$
Correct Answer: $\tfrac{7}{2}R$
Explanation: A diatomic molecule has 5 degrees of freedom (3 translational + 2 rotational). Internal energy = $\tfrac{5}{2}RT$, so $C_v = \tfrac{5}{2}R$. Using relation $C_p = C_v + R$, we get $C_p = \tfrac{7}{2}R$.
183. The difference between molar heat capacities at constant pressure and volume is:
ⓐ. $C_p – C_v = \tfrac{R}{2}$
ⓑ. $C_p – C_v = R$
ⓒ. $C_p – C_v = 2R$
ⓓ. $C_p – C_v = 0$
Correct Answer: $C_p – C_v = R$
Explanation: Thermodynamics gives the Mayer’s relation: $C_p – C_v = R$. This is independent of gas type (ideal gases) and directly connects microscopic energy distribution with macroscopic heat processes.
184. The adiabatic index $\gamma$ for a monatomic gas is:
ⓐ. $\tfrac{5}{3}$
ⓑ. $\tfrac{7}{5}$
ⓒ. $\tfrac{3}{2}$
ⓓ. $2$
Correct Answer: $\tfrac{5}{3}$
Explanation: For monatomic gases, $C_p = \tfrac{5}{2}R$, $C_v = \tfrac{3}{2}R$. Thus, $\gamma = \tfrac{C_p}{C_v} = \tfrac{5/2}{3/2} = \tfrac{5}{3}$. This ratio appears in adiabatic process equations.
185. For a diatomic gas (no vibrations), the adiabatic index $\gamma$ is:
ⓐ. $\tfrac{5}{3}$
ⓑ. $\tfrac{7}{5}$
ⓒ. $\tfrac{9}{7}$
ⓓ. $2$
Correct Answer: $\tfrac{7}{5}$
Explanation: For diatomic molecules, $C_v = \tfrac{5}{2}R$, $C_p = \tfrac{7}{2}R$. So, $\gamma = \tfrac{C_p}{C_v} = \tfrac{7/2}{5/2} = \tfrac{7}{5}$.
186. The relation between pressure and volume in an adiabatic process is:
ⓐ. $PV = \text{constant}$
ⓑ. $PV^\gamma = \text{constant}$
ⓒ. $\tfrac{V}{T} = \text{constant}$
ⓓ. $\tfrac{P}{T} = \text{constant}$
Correct Answer: $PV^\gamma = \text{constant}$
Explanation: From thermodynamics, for an adiabatic process $dQ = 0$. Using first law and ideal gas relations, we obtain $PV^\gamma = \text{constant}$. This law is used to describe compression and expansion in engines.
187. The work done by an ideal gas in an isothermal process from $V_1$ to $V_2$ is:
ⓐ. $W = P \Delta V$
ⓑ. $W = nR \Delta T$
ⓒ. $W = nRT \ln \tfrac{V_2}{V_1}$
ⓓ. $W = \tfrac{3}{2}nR \Delta T$
Correct Answer: $W = nRT \ln \tfrac{V_2}{V_1}$
Explanation: In an isothermal process, $\Delta U = 0$. From the first law, $Q = W$. Work done is $W = \int_{V_1}^{V_2} P dV = nRT \ln \tfrac{V_2}{V_1}$.
188. In an adiabatic expansion of 2 moles of a monatomic gas from 5 L to 20 L at 300 K, the final temperature is approximately:
ⓐ. 100 K
ⓑ. 150 K
ⓒ. 200 K
ⓓ. 250 K
Correct Answer: 200 K
Explanation: For adiabatic process: $TV^{\gamma-1} = \text{constant}$. For monatomic gas, $\gamma = 5/3$. So, $T_2 = T_1 \left( \tfrac{V_1}{V_2} \right)^{\gamma-1} = 300 \times (5/20)^{2/3} = 200 \, K$.
189. The specific heat at constant volume of 1 mole of an ideal monatomic gas is:
ⓐ. $\tfrac{3}{2}R$ J/mol·K
ⓑ. $\tfrac{5}{2}R$ J/mol·K
ⓒ. $R$ J/mol·K
ⓓ. $\tfrac{7}{2}R$ J/mol·K
Correct Answer: $\tfrac{3}{2}R$ J/mol·K
Explanation: From equipartition, 3 translational degrees of freedom yield internal energy $U = \tfrac{3}{2}RT$. Thus, $C_v = \tfrac{dU}{dT} = \tfrac{3}{2}R$.
190. Which thermodynamic process involves heat transfer but no change in temperature?
ⓐ. Isothermal process
ⓑ. Adiabatic process
ⓒ. Isobaric phase transition (melting/boiling)
ⓓ. Isochoric process
Correct Answer: Isobaric phase transition (melting/boiling)
Explanation: During melting or boiling, heat energy supplied is used to change phase (latent heat) without changing temperature. This cannot be explained by ideal gas laws alone but connects to thermodynamics and specific heat concepts.
191. The molar specific heat at constant volume $C_v$ is defined as:
ⓐ. Heat required to raise the temperature of 1 kg substance by 1°C at constant pressure
ⓑ. Heat required to raise the temperature of 1 mole gas by 1 K at constant volume
ⓒ. Heat required to melt 1 mole of substance at constant pressure
ⓓ. Heat required to vaporize 1 mole of liquid at constant volume
Correct Answer: Heat required to raise the temperature of 1 mole gas by 1 K at constant volume
Explanation: By definition, $C_v = \left(\frac{dQ}{ndT}\right)_V$. Since volume is constant, no work is done, and heat goes entirely into raising internal energy.
192. For an ideal gas, the relation between $C_p$ and $C_v$ is given by:
ⓐ. $C_p = C_v$
ⓑ. $C_p – C_v = R$
ⓒ. $C_p + C_v = R$
ⓓ. $C_p = 2C_v$
Correct Answer: $C_p – C_v = R$
Explanation: Mayer’s relation shows the difference in specific heats equals the universal gas constant $R$. This comes from the fact that at constant pressure, the system does additional work against the external pressure.
193. For a monatomic ideal gas, what is the value of $C_p$?
ⓐ. $R$
ⓑ. $\tfrac{3}{2}R$
ⓒ. $\tfrac{5}{2}R$
ⓓ. $3R$
Correct Answer: $\tfrac{5}{2}R$
Explanation: For monatomic gases, $C_v = \tfrac{3}{2}R$. Using Mayer’s relation, $C_p = C_v + R = \tfrac{5}{2}R$. This matches experimental data for noble gases like helium and neon.
194. For a diatomic gas (ignoring vibrational energy), the molar heat capacity at constant volume is:
ⓐ. $\tfrac{3}{2}R$
ⓑ. $\tfrac{5}{2}R$
ⓒ. $\tfrac{7}{2}R$
ⓓ. $4R$
Correct Answer: $\tfrac{5}{2}R$
Explanation: A diatomic molecule has 5 degrees of freedom. Hence, internal energy = $\tfrac{5}{2}RT$, so $C_v = \tfrac{dU}{dT} = \tfrac{5}{2}R$.
195. Calculate the molar heat capacity ratio ($\gamma = C_p / C_v$) for a monatomic ideal gas.
ⓐ. 1.33
ⓑ. 1.40
ⓒ. 1.50
ⓓ. 1.67
Correct Answer: 1.67
Explanation: For a monatomic gas, $C_v = \tfrac{3}{2}R$ and $C_p = \tfrac{5}{2}R$. Hence, $\gamma = \tfrac{C_p}{C_v} = \tfrac{5/2}{3/2} = 1.67$.
196. A diatomic ideal gas at room temperature has $C_v = \tfrac{5}{2}R$. What is its $C_p$?
ⓐ. $\tfrac{5}{2}R$
ⓑ. $\tfrac{7}{2}R$
ⓒ. $3R$
ⓓ. $4R$
Correct Answer: $\tfrac{7}{2}R$
Explanation: Using Mayer’s relation, $C_p = C_v + R = \tfrac{5}{2}R + R = \tfrac{7}{2}R$. Thus, $\gamma = \tfrac{7}{5} = 1.40$.
197. The specific heat at constant pressure is greater than at constant volume because:
ⓐ. At constant pressure, gas does external work in addition to raising internal energy.
ⓑ. At constant volume, gas has more degrees of freedom.
ⓒ. At constant pressure, intermolecular forces vanish.
ⓓ. At constant volume, energy is destroyed.
Correct Answer: At constant pressure, gas does external work in addition to raising internal energy.
Explanation: Heat supplied at constant pressure is used both for increasing internal energy and for expansion work, hence $C_p > C_v$.
198. For 2 moles of a monatomic ideal gas, calculate the increase in internal energy when temperature increases by 50 K.
ⓐ. $50R$
ⓑ. $75R$
ⓒ. $100R$
ⓓ. $150R$
Correct Answer: $150R$
Explanation: $\Delta U = nC_v\Delta T = 2 \times \tfrac{3}{2}R \times 50 = 150R$.
199. The heat capacity ratio $\gamma$ is important in which thermodynamic process?
ⓐ. Isothermal expansion
ⓑ. Isochoric heating
ⓒ. Adiabatic expansion or compression
ⓓ. Phase transitions
Correct Answer: Adiabatic expansion or compression
Explanation: For adiabatic processes, the relation $PV^\gamma = \text{constant}$ holds, where $\gamma = \tfrac{C_p}{C_v}$. This is critical in understanding sound waves and engine cycles.
200. For a gas mixture containing 1 mole of monatomic gas and 1 mole of diatomic gas (no vibrations), the effective molar $C_v$ is:
ⓐ. $2R$
ⓑ. $\tfrac{3}{2}R$
ⓒ. $\tfrac{4}{2}R$
ⓓ. $\tfrac{8}{2}R$
Correct Answer: $2R$
Explanation: Monatomic gas: $C_v = \tfrac{3}{2}R$. Diatomic gas: $C_v = \tfrac{5}{2}R$. For 1 mole each, average $C_v = \tfrac{(1.5R + 2.5R)}{2} \times 2 = 2R$. Thus, effective $C_v = 2R$.
The chapter Kinetic Theory in Class 11 Physics (NCERT/CBSE syllabus) is essential for mastering the microscopic understanding of gases and their thermodynamic behavior. It covers a wide range of concepts such as ideal gas equation, kinetic interpretation of temperature, pressure of a gas based on molecular motion, and the distribution of molecular speeds. The chapter also explains Brownian motion, mean free path, molecular collisions, and Avogadro’s hypothesis. Questions from this chapter frequently appear in board exams and in competitive exams like JEE and NEET, testing both theory and numerical problem-solving skills. Out of the total 498 MCQs with solutions, this part includes the **second set of 100 questions** with step-by-step explanations to improve your accuracy and exam readiness.
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