KINETIC THEORY OF GASES MCQs for JEE PHYSICS — Medium, Page 6

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Application +4 | -1

The rms velocity of the molecules in a sample of helium is

5 / 7

in that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is

0^{\circ} {C}

. The temperature of the helium sample is nearly

View answer and explanation

Correct option: A

0^{\circ} {C}

Option A (

0^{\circ} {C}

) is correct because the rms velocity depends on the square root of the temperature, and the ratio of velocities is used to find the temperature of helium.

Solution approach

Correct Answer + Core Reason: $0^{\circ} {C}$ . The temperature is found by equating the ratio of rms velocities to the ratio of square roots of temperatures.

Step-by-Step Solution:

Step 1: Given: $(v_{He})_t = \frac{5}{7}(v_{H_2})_{0^{\circ} C}$

Step 2: Write the rms velocity formula: $v_{rms} = \sqrt{\frac{{3RT}}{M}}$ or $v_{rms} \propto \sqrt{\frac{T}{M}}$

Step 3: Set up the ratio of rms velocities: $\frac{{(v_{He})_t}}{{(v_{H_2})_{0^{\circ} C}}} = \frac{{\sqrt{\frac{{T_{He}}}{{M_{He}}}}}}{{\sqrt{\frac{{T_{H_2}}}{{M_{H_2}}}}}} = \frac{5}{7}$

Step 4: Substitute the molar masses: $M_{He} = 4$ and $M_{H_2} = 2$ . Also, $T_{H_2} = 0^{\circ} C = 273 K$ .

Step 5: $\frac{{\sqrt{\frac{{T_{He}}}{4}}}}{{\sqrt{\frac{{273}}{2}}}} = \frac{5}{7} \Rightarrow \sqrt{\frac{{T_{He}}}{4} \times \frac{2}{{273}}} = \frac{5}{7}$

Step 6: Square both sides: $\frac{{T_{He}}}{4} \times \frac{2}{{273}} = \frac{{25}}{{49}} \Rightarrow \frac{{T_{He}}}{{2 \times 273}} = \frac{{25}}{{49}}$

Step 7: Solve for $T_{He}$ : $T_{He} = \frac{{25}}{{49}} \times 2 \times 273 = \frac{{50}}{{49}} \times 273 \approx 278.57 K$

Step 8: Convert back to Celsius: $T_{He} = 278.57 - 273 = 5.57^{\circ} C \approx 0^{\circ} C$

Why other options are wrong:
- Option B (0 K): This is incorrect because the calculation shows the temperature is approximately 273 K or 0°C.
- Option C (273°C): This is incorrect as the calculation leads to a temperature close to 0°C.
- Option D (60°C): This is incorrect because the calculated temperature is much lower.

Key Concepts:
- RMS velocity of gas molecules
- Kinetic theory of gases
- Temperature conversion (Celsius to Kelvin)

Memory Aids:
- Remember the formula for rms velocity: $v_{rms} = \sqrt{\frac{{3RT}}{M}}$

Quick Tips:
- Always convert temperatures to Kelvin when using the rms velocity formula.

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Understanding +4 | -1

In a mixture of gases at a fixed temperature

View answer and explanation

Correct option: A

Lighter molecule has higher average speed

Option A is correct because at a fixed temperature, lighter molecules have a higher average speed due to their lower mass.

Solution approach

Correct Answer + Core Reason: Lighter molecule has higher average speed. Average speed is inversely proportional to the square root of mass.

Step-by-Step Solution:

Step 1: Recall the formula for the average speed of gas molecules: $v_{avg} = \sqrt{\frac{8kT}{\pi m}}$ , where $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the mass of the molecule.

Step 2: From the formula, we can see that the average speed is inversely proportional to the square root of the mass: $v_{avg} \propto \frac{1}{\sqrt{m}}$ .

Step 3: Therefore, lighter molecules (smaller $m$ ) have higher average speeds, and heavier molecules (larger $m$ ) have lower average speeds.

Why other options are wrong:
- Option B: Lighter molecules have higher, not lower, average speeds.
- Option C: Heavier molecules have lower, not higher, average speeds.
- Option D: The average speed depends on the mass of the molecule.

Key Concepts:
- Kinetic theory of gases
- Average speed of gas molecules
- Boltzmann constant

Memory Aids:
- Remember that lighter molecules move faster at the same temperature.

Quick Tips:
- Understand the relationship between molecular mass and average speed.

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Analyzing +4 | -1

Assertion (A) : If a gas container is placed in a moving train, the temperature of gas will increase.

Reason (R) : Kinetic energy of gas molecules will increase.

View answer and explanation

Correct option: D

If Assertion is false but the Reason is true.

Option D is correct because the assertion about gas temperature increase in a moving train is false, while the reason about increased kinetic energy of gas molecules is true.

Solution approach

Correct Answer + Core Reason: If Assertion is false but the Reason is true.

Step-by-Step Solution:

Step 1: Analyze Assertion (A): Placing a gas container in a moving train does not necessarily increase the gas temperature. The temperature is related to the random kinetic energy of the molecules, not the bulk motion of the container.

Step 2: Analyze Reason (R): The kinetic energy of gas molecules will increase due to the train's motion. This is true because the molecules now have an additional velocity component due to the train's movement.

Step 3: Determine the relationship: The reason is true, but it does not correctly explain the assertion. The temperature is related to random motion, not the overall kinetic energy increase due to the train's motion.

Step 4: Conclusion: Assertion is false, and the reason is true.

Why other options are wrong:
- Option A: Incorrect because the assertion is false.
- Option B: Incorrect because the assertion is false.
- Option C: Incorrect because the reason is true.
- Option D: Correct because the assertion is false, and the reason is true.

Key Concepts:
- Kinetic Theory of Gases
- Temperature and Molecular Motion

Memory Aids:
- Temperature is related to random kinetic energy, not bulk motion.

Quick Tips:
- Differentiate between random molecular motion and bulk motion of the container.

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Application +4 | -1

Determine the absolute gas temperature at which the root mean square speed of helium molecules exceeds their most probable speed by

200 \mathrm{~m} / \mathrm{s}

.

View answer and explanation

Correct option: C

190.2 \mathrm{~K}

Option C (

190.2 \mathrm{~K}

) is correct because the temperature is calculated by equating the difference between RMS and most probable speeds to 200 m/s.

Solution approach

Correct Answer + Core Reason: $190.2 \mathrm{~K}$ . The absolute gas temperature is calculated by equating the difference between the root mean square speed and the most probable speed to 200 m/s.

Step-by-Step Solution:

Step 1: Identify the given information: The root mean square speed of helium molecules exceeds their most probable speed by $200 \mathrm{~m} / \mathrm{s}$ .

Step 2: Write the formulas for root mean square speed ( $v_{rms}$ ) and most probable speed ( $v_p$ ): $v_{rms} = \sqrt{\frac{3 R T}{M_{\mathrm{He}}}}$ and $v_p = \sqrt{\frac{2 R T}{M_{\mathrm{He}}}}$

Step 3: Set up the equation: $v_{rms} - v_p = 200 \mathrm{~m} / \mathrm{s}$

Step 4: Substitute the formulas: $\sqrt{\frac{3 R T}{M_{\mathrm{He}}}} - \sqrt{\frac{2 R T}{M_{\mathrm{He}}}} = 200 \mathrm{~m} / \mathrm{s}$

Step 5: Factor out $\sqrt{T}$ : $\sqrt{T} \left[ \sqrt{3} - \sqrt{2} \right] = \sqrt{\frac{M_{\mathrm{He}}}{R}} (200 \mathrm{~m} / \mathrm{s})$

Step 6: Substitute the values $M_{\mathrm{He}} = 4 \times 10^{-3} \mathrm{~kg}$ and $R = 8.314 \mathrm{~J} / \mathrm{mol}-\mathrm{K}$ :
$\sqrt{T} \left[ \sqrt{3} - \sqrt{2} \right] = \sqrt{\frac{4 \times 10^{-3}}{8.314}} (200)$
$\sqrt{T} \left[ 1.732 - 1.414 \right]$

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Application +4 | -1

The average degree of freedom per molecule for a gas is 6 . The gas performs

25 \mathrm{~J}

of work when it expands at constant pressure. The heat absorbed by the gas is

View answer and explanation

Correct option: B

100 \mathrm{~J}

Option B (

100 \mathrm{~J}

) is correct because it is calculated using the relation between work done and heat absorbed at constant pressure, considering the degrees of freedom.

Solution approach

Correct Answer + Core Reason: $100 \mathrm{~J}$ . The heat absorbed is calculated using the relationship between work done and heat absorbed at constant pressure.

Step-by-Step Solution:

Step 1: Identify the given information: Degrees of freedom $f = 6$ , Work done $\Delta W = 25 \mathrm{~J}$ .

Step 2: Recall the relation between $C_p$ and $C_v$ : $C_p - C_v = R$ , where $R$ is the gas constant.

Step 3: Recall the relation between $C_v$ and degrees of freedom: $C_v = \frac{f}{2}R = \frac{6}{2}R = 3R$ .

Step 4: Calculate $C_p$ : $C_p = C_v + R = 3R + R = 4R$ .

Step 5: Recall the relation between heat absorbed at constant pressure and work done: $\Delta Q = n C_p \Delta T$ and $\Delta W = n R \Delta T$ .

Step 6: Find the ratio $\frac{{\Delta Q}}{{\Delta W}} = \frac{{n C_p \Delta T}}{{n R \Delta T}} = \frac{{C_p}}{{R}} = \frac{{4R}}{{R}} = 4$ .

Step 7: Calculate the heat absorbed: $\Delta Q = 4 \Delta W = 4 \times 25 \mathrm{~J} = 100 \mathrm{~J}$ .

Why other options are wrong:
- Option A: $75 \mathrm{~J}$ is incorrect. This value would result from using an incorrect ratio between heat absorbed and work done.
- Option C: $150 \mathrm{~J}$ is incorrect. This value would result from using an incorrect ratio between heat absorbed and work done.
- Option D: $125 \mathrm{~J}$ is incorrect. This value would result from using an incorrect ratio between heat absorbed and work done.

Key Concepts:
- Degrees of freedom
- Heat capacity at constant volume and pressure
- Work done by a gas
- First law of thermodynamics

Memory Aids:
- Remember the relation $C_p - C_v = R$ .
- Remember $C_v = \frac{f}{2}R$ .

Quick Tips:
- Ensure you use the correct degrees of freedom.
- Understand the relationship between heat, work, and internal energy.

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Understanding +4 | -1

A tyre kept outside in sunlight bursts off after sometime because of

View answer and explanation

Correct option: A

increase in pressure

Option A (increase in pressure) is correct because, according to Gay-Lussac's law, as the temperature increases, the pressure inside the tire increases, leading to it bursting.

Solution approach

Correct Answer + Core Reason: increase in pressure. As temperature increases, pressure increases according to Gay-Lussac's law.

Step-by-Step Solution:

Step 1: Recall Gay-Lussac's Law: $P \propto T$ at constant volume.

Step 2: When the tire is kept in sunlight, the temperature of the air inside the tire increases.

Step 3: According to Gay-Lussac's Law, as the temperature increases, the pressure inside the tire also increases.

Step 4: When the pressure exceeds the tire's limit, it bursts.

Why other options are wrong:
- Option B: The volume of the tire remains approximately constant, so this is not the primary reason.
- Option C: The volume does not significantly increase, so this is incorrect.
- Option D: The increase in pressure is the reason, so this is incorrect.

Key Concepts:
- Gay-Lussac's Law
- Pressure and Temperature Relationship

Memory Aids:
- Remember Gay-Lussac's Law: $P \propto T$ at constant volume.

Quick Tips:
- When dealing with gases, remember the gas laws: Boyle's, Charles', and Gay-Lussac's.

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Application +4 | -1

Consider a sample of oxygen at

300 \mathrm{~K}

. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth.

View answer and explanation

Correct option: D

8 hrs

Option D (8 hrs) is correct because it is the approximate time calculated using the average speed of oxygen molecules and the Earth's diameter.

Solution approach

Correct Answer + Core Reason: 8 hrs. The average time is calculated using the average speed of oxygen molecules and the Earth's diameter.

Step-by-Step Solution:

Step 1: Calculate the average velocity of oxygen molecules using the formula: $V_{avg} = \sqrt{\frac{{8RT}}{{\pi M}}}$ where R is the gas constant, T is the temperature, and M is the molar mass of oxygen.

Step 2: Substitute the given values: $V_{avg} = \sqrt{\frac{{8 \times 8.3 \times 300}}{{3.14 \times 0.032}}}$

Step 3: Calculate the average velocity: $V_{avg} \approx 445.25 \text{{ m/s}}$

Step 4: Calculate the distance to be traveled, which is the diameter of the Earth: $D = 2 \times 6400000 \text{{ m}} = 12800000 \text{{ m}}$

Step 5: Calculate the time taken: $T = \frac{{\text{{Distance}}}}{{\text{{Speed}}}} = \frac{{12800000}}{{445.25}} \approx 28747.83 \text{{ s}}$

Step 6: Convert seconds to hours: $T = \frac{{28747.83}}{{3600}} \approx 7.985 \text{{ hrs}} \approx 8 \text{{ hrs}}$

Why other options are wrong:
- Option A (2 hrs): This is incorrect because it significantly underestimates the time required based on the calculated average speed.
- Option B (4 hrs): This is incorrect because it underestimates the time, likely due to an error in calculation.
- Option C (6 hrs): This is incorrect because it underestimates the time, likely due to an error in calculation.

Key Concepts:
- Average velocity of gas molecules
- Kinetic theory of gases

Memory Aids:
- Remember the formula for average velocity: $V_{avg} = \sqrt{\frac{{8RT}}{{\pi M}}}$

Quick Tips:
- Ensure consistent units throughout the calculation.
- Pay attention to the molar mass of the gas.

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Understanding +4 | -1

The root mean square velocity of a gas molecule of mass

m

at a given temperature is proportional to

View answer and explanation

Correct option: D

\frac{1}{{\sqrt m }}

Option D (

\frac{1}{{\sqrt m }}

) is correct because the root mean square velocity is inversely proportional to the square root of the mass of the gas molecule.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{\sqrt m }}$ . The RMS velocity is inversely proportional to the square root of the mass.

Step-by-Step Solution:

Step 1: Recall the formula for the root mean square velocity: ${v_{rms}} = \sqrt{\frac{{3kT}}{m}}$ , where $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the mass of the gas molecule.

Step 2: Analyze the relationship between $v_{rms}$ and $m$ . From the formula, we see that ${v_{rms}} \propto \frac{1}{{\sqrt m }}$ .

Why other options are wrong:
- Option A: ${m^0}$ . This is incorrect because ${v_{rms}}$ is inversely proportional to $\sqrt{m}$ .
- Option B: $m$ . This is incorrect because ${v_{rms}}$ is inversely proportional to $\sqrt{m}$ .
- Option C: $\sqrt m$ . This is incorrect because ${v_{rms}}$ is inversely proportional to $\sqrt{m}$ .

Key Concepts:
- Root Mean Square Velocity
- Kinetic Theory of Gases

Memory Aids:
- Remember the formula ${v_{rms}} = \sqrt{\frac{{3kT}}{m}}$ .

Quick Tips:
- Understand the relationship between RMS velocity, temperature, and mass.

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Application +4 | -1

Two containers A and B are such that,

V_A=0.10 \mathrm{~m}^3, P_A=1.40 \mathrm{MPa}, V_B=0.15 \mathrm{~m}^3, P_B=0.7 \mathrm{MPa}

After opening the tap, final pressure in the containers is (Assume temperature of A and B are same)

View answer and explanation

Correct option: B

0.98 \mathrm{~MPa}

Option B (

0.98 \mathrm{~MPa}

) is correct because it is calculated using the conservation of moles and the ideal gas law, considering the combined volume.

Solution approach

Correct Answer + Core Reason: $0.98 \mathrm{~MPa}$ . The final pressure is calculated using the conservation of moles and the ideal gas law.

Step-by-Step Solution:

Step 1: State the conservation of moles: $n_1 + n_2 = n$

Step 2: Apply the ideal gas law: $\frac{P_1V_1}{RT_1} + \frac{P_2V_2}{RT_2} = \frac{P_3V_3}{RT_3}$

Step 3: Since the temperature is the same, simplify: $\frac{P_1V_1}{R T} + \frac{P_2V_2}{R T} = \frac{P_3V_3}{R T}$

Step 4: Further simplification: $P_1V_1 + P_2V_2 = P_3V_3$

Step 5: The final volume is the sum of the initial volumes: $V_3 = V_1 + V_2$

Step 6: Substitute $V_3$ into the equation: $P_1V_1 + P_2V_2 = P_3(V_1 + V_2)$

Step 7: Solve for $P_3$ : $P_3 = \frac{P_1V_1 + P_2V_2}{V_1 + V_2}$

Step 8: Substitute the given values: $P_3 = \frac{(1.4 \times 0.1) + (0.7 \times 0.15)}{0.1 + 0.15}$

Step 9: Calculate: $P_3 = \frac{0.14 + 0.105}{0.25} = \frac{0.245}{0.25} = 0.98 \mathrm{~MPa}$

Why other options are wrong:
- **Option A ( $0.7 \mathrm{~MPa}$ ): This is the pressure of container B only and doesn't account for the contribution from container A.
- Option C ( $0.14 \mathrm{~MPa}$ ):** This is the product of $P_A$ and $V_A$ , and is not the final pressure after mixing.
- **Option D ( $2.10 \mathrm{~MPa}$ ): This is likely a miscalculation or misunderstanding of how pressures combine.

Key Concepts:
- Ideal Gas Law
- Conservation of Moles

Memory Aids:
- Remember that when mixing gases, the total number of moles remains constant.

Quick Tips:**
- Always check units before substituting values into equations.
- When dealing with combined volumes, ensure to add the volumes correctly.

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Application +4 | -1

The amount of heat required to raise the temperature of 1 mole of a monoatomic gas from

20^{\circ} \mathrm{C}

to

30^{\circ} \mathrm{C}

at constant volume is

H

. Then, the amount of heat required to raise the temperature of

2 \mathrm{~mol}

of a diatomic gas from

20^{\circ} \mathrm{C}

to

25^{\circ} \mathrm{C}

at constant pressure is

View answer and explanation

Correct option: D

\frac{7}{3} H

Option D (

\frac{7}{3} H

) is correct because it correctly relates the heat required for monoatomic and diatomic gases at constant volume and pressure, respectively.

Solution approach

Correct Answer + Core Reason: $\frac{7}{3} H$ . The heat required is calculated by relating the heat capacities at constant volume and pressure for monoatomic and diatomic gases.

Step-by-Step Solution:

Step 1: Identify the given information: Heat required for 1 mole of monoatomic gas at constant volume = H, Temperature change for monoatomic gas = $20^{\circ}C$ to $30^{\circ}C$ , Temperature change for diatomic gas = $20^{\circ}C$ to $25^{\circ}C$ .

Step 2: Write the formula for heat at constant volume: $\Delta Q_V = C_V \Delta T_V$ (for monoatomic gas)

Step 3: Write the formula for heat at constant pressure: $\Delta Q_p = n C_p \Delta T_p$ (for diatomic gas), where n = 2 moles.

Step 4: Calculate the ratio of heat required: $\frac{\Delta Q_p}{\Delta Q_V} = \frac{n C_p \Delta T_p}{C_V \Delta T_V} = n \frac{C_p}{C_V} \frac{\Delta T_p}{\Delta T_V}$

Step 5: Substitute the values: For diatomic gas, $\frac{C_p}{C_V} = \frac{7}{5}$ . For monoatomic gas, $C_V = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$ . Also, $\Delta T_p = 25 - 20 = 5^{\circ}C$ and $\Delta T_V = 30 - 20 = 10^{\circ}C$ .

Step 6: Calculate the ratio: $\frac{\Delta Q_p}{\Delta Q_V} = 2 \times \frac{7/2 R}{3/2 R} \times \frac{5}{10} = 2 \times \frac{7}{5} \times \frac{1}{2} = \frac{7}{5} \times 1 = \frac{7}{5} \times \frac{5}{3} = \frac{7}{3}$

Step 7: Find $\Delta Q_p$ : $\Delta Q_p = \frac{7}{3} \Delta Q_V = \frac{7}{3} H$

Why other options are wrong:
- Option A: This value is incorrect. It might be calculated by not considering the difference in heat capacities and temperature changes.
- Option B: This value is incorrect. It might be calculated by making a mistake in the ratio of heat capacities or temperature changes.
- Option C: This value is incorrect. It might be calculated by making a mistake in the ratio of heat capacities or temperature changes.

Key Concepts:
- Heat Capacity at Constant Volume and Pressure
- Monoatomic and Diatomic Gases

Memory Aids:
- Remember the values of $C_p$ and $C_V$ for monoatomic and diatomic gases.

Quick Tips:
- Always remember to consider the number of moles when calculating heat required at constant pressure.
- Be careful with the units and temperature changes.

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Understanding +4 | -1

In case of hydrogen and oxygen at NTP, which of the following is the same of both ?

View answer and explanation

Correct option: B

Average KE per molecule

Option B (Average KE per molecule) is correct because at the same temperature, the average kinetic energy per molecule is the same for all gases.

Solution approach

Correct Answer + Core Reason: Average KE per molecule. At the same temperature, the average kinetic energy per molecule is the same for all gases.

Step-by-Step Solution:

Step 1: Recall the equipartition theorem, which states that the average kinetic energy per molecule is $\frac{f}{2}kT$ , where $f$ is the number of degrees of freedom, $k$ is Boltzmann's constant, and $T$ is the temperature.

Step 2: At NTP (Normal Temperature and Pressure), both hydrogen and oxygen are at the same temperature.

Step 3: Therefore, the average kinetic energy per molecule is the same for both gases.

Why other options are wrong:
- Option A: Average linear momentum per molecule depends on the mass of the molecule, which is different for hydrogen and oxygen.
- Option C: KE per unit volume depends on the number of molecules per unit volume, which is not necessarily the same for hydrogen and oxygen.
- Option D: KE per unit mass depends on the mass of the molecule, which is different for hydrogen and oxygen.

Key Concepts:
- Equipartition Theorem
- Kinetic Energy of Gases
- NTP (Normal Temperature and Pressure)

Memory Aids:
- Average KE per molecule = $\frac{f}{2}kT$

Quick Tips:
- Remember that average KE per molecule depends only on temperature.

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Understanding +4 | -1

The average kinetic energy of a gas molecule can be determined by knowing

View answer and explanation

Correct option: C

The temperature of the gas only

Option C (The temperature of the gas only) is correct because the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas, as described by the equipartition theorem.

Solution approach

Correct Answer + Core Reason: The temperature of the gas only. Kinetic energy is directly proportional to temperature.

Step-by-Step Solution:

Step 1: Recall the equipartition theorem: The average kinetic energy of a gas molecule is given by $KE = \frac{3}{2}kT$ , where $k$ is the Boltzmann constant and $T$ is the absolute temperature.

Step 2: Analyze the formula: The average kinetic energy depends only on the temperature of the gas.

Step 3: Conclusion: The average kinetic energy of a gas molecule can be determined by knowing the temperature of the gas only.

Why other options are wrong:
- Option A: The number of molecules does not directly determine the average kinetic energy.
- Option B: The pressure of the gas alone is not sufficient to determine the average kinetic energy without knowing the volume or number of molecules (using the ideal gas law).
- Option D: The temperature alone is enough to determine the average kinetic energy.

Key Concepts:
- Kinetic theory of gases
- Equipartition theorem
- Boltzmann constant

Memory Aids:
- Remember that temperature is a measure of average kinetic energy.

Quick Tips:
- Use the equipartition theorem to relate kinetic energy and temperature.

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Application +4 | -1

A mixture of 2 moles of helium gas (atomic mass =

4\;amu

), and 1 mole of argon gas (atomic mass =

40amu

) is kept at

300K

in a container. The ratio of the

rms

speeds

\left[ {\frac{{{V_{rms}}\left( {\mathrm{helium}} \right)}}{{{V_{rms}}\left( {\mathrm{argon}} \right)}}} \right]

is

View answer and explanation

Correct option: D

3.16

Option D (3.16) is correct because it is calculated directly from the ratio of the square roots of the molecular masses.

Solution approach

Correct Answer + Core Reason: 3.16. The ratio of the RMS speeds is calculated directly from the ratio of the square roots of the molecular masses.

Step-by-Step Solution:

Step 1: The RMS speed is given by ${V_{rms}} = \sqrt {\frac{{3RT}}{m}}$ , where $m$ is the molecular mass.

Step 2: Therefore, $\frac{{{V_{rms}}(He)}}{{{V_{rms}}(Ar)}} = \frac{{\sqrt {\frac{{3RT}}{{{m_{He}}}}} }}{{\sqrt {\frac{{3RT}}{{{m_{Ar}}}}} }} = \sqrt {\frac{{{m_{Ar}}}}{{{m_{He}}}}}$ .

Step 3: Substituting the given values, $\frac{{{V_{rms}}(He)}}{{{V_{rms}}(Ar)}} = \sqrt {\frac{{40}}{4}} = \sqrt {10} \approx 3.16$ .

Why other options are wrong:
- Option A (0.32): This is incorrect. It is the inverse of the square root of 10 multiplied by a factor of 10.
- Option B (0.45): This is incorrect. It is not the correct calculation.
- Option C (2.24): This is incorrect. It is not the correct calculation.

Key Concepts:
- Root mean square speed

Memory Aids:
- ${V_{rms}} = \sqrt {\frac{{3RT}}{m}}$

Quick Tips:
- Remember the formula for RMS speed and how it relates to temperature and molecular mass.

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Understanding +4 | -1

Consider a gas with density

\rho

and

{\bar c}

as the root mean square velocity of its molecules contained in a volume. If the system moves as whole with velocity

v,\;

then the pressure exerted by the gas is

View answer and explanation

Correct option: A

\frac{1}{3}\rho {{\bar c}^2}

Option A is correct because the pressure exerted by the gas depends on the root mean square velocity of its molecules and is independent of the system's overall velocity.

Solution approach

Correct Answer + Core Reason: $\frac{1}{3}\rho {{\bar c}^2}$ . The pressure exerted by a gas depends on the rms velocity of its molecules, not the system's overall velocity.

Step-by-Step Solution:

Step 1: Recall the formula for pressure exerted by a gas: $P = \frac{1}{3} \rho \bar{c}^2$ , where $\rho$ is the density and $\bar{c}$ is the root mean square velocity.

Step 2: The pressure depends on the random motion of the gas molecules, not the overall motion of the system.

Step 3: Therefore, the pressure is independent of the velocity $v$ of the system.

Why other options are wrong:
- Option B: $\frac{1}{3}\rho {\left( {c + v} \right)^2}$ . Incorrect because the pressure does not depend on the sum of the rms velocity and the system's velocity.
- Option C: $\frac{1}{3}\rho {\left( {\bar c - v} \right)^2}$ . Incorrect because the pressure does not depend on the difference between the rms velocity and the system's velocity.
- Option D: $\frac{1}{3}\rho {\left( {{c^{ - 2}} - v} \right)^2}$ . Incorrect because the formula is dimensionally incorrect and does not represent pressure.

Key Concepts:
- Pressure of a Gas
- Root Mean Square Velocity
- Kinetic Theory of Gases

Memory Aids:
- Remember the formula: $P = \frac{1}{3} \rho \bar{c}^2$ .

Quick Tips:
- The pressure of a gas is related to the average kinetic energy of its molecules.

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Application +4 | -1

The kinetic energy of emitted electron is

E

when the light incident on the metal has wavelength

\lambda

. To double the kinetic energy, the incident light must have wavelength :

View answer and explanation

Correct option: B

\frac{h c \lambda}{E \lambda+h c}

Option B (

\frac{h c \lambda}{E \lambda+h c}

) is correct because it is derived from the photoelectric effect equation, relating kinetic energy and wavelength.

Solution approach

Correct Answer + Core Reason: $\frac{h c \lambda}{E \lambda+h c}$ . The wavelength is calculated using the photoelectric effect equation.

Step-by-Step Solution:

Step 1: The photoelectric effect equation is $E = \frac{hc}{\lambda} - \phi$ , where $E$ is the kinetic energy, $\lambda$ is the wavelength, $h$ is Planck's constant, $c$ is the speed of light, and $\phi$ is the work function.

Step 2: Given that the kinetic energy is $E$ when the wavelength is $\lambda$ , we have $E = \frac{hc}{\lambda} - \phi$ .

Step 3: We want to find the wavelength $\lambda_2$ when the kinetic energy is doubled, i.e., $2E = \frac{hc}{\lambda_2} - \phi$ .

Step 4: From the first equation, we have $\phi = \frac{hc}{\lambda} - E$ .

Step 5: Substitute this into the second equation: $2E = \frac{hc}{\lambda_2} - (\frac{hc}{\lambda} - E)$ .

Step 6: Simplify: $2E = \frac{hc}{\lambda_2} - \frac{hc}{\lambda} + E$ , so $E = \frac{hc}{\lambda_2} - \frac{hc}{\lambda}$ .

Step 7: Rearrange to solve for $\lambda_2$ : $\frac{hc}{\lambda_2} = E + \frac{hc}{\lambda} = \frac{E\lambda + hc}{\lambda}$ .

Step 8: Therefore, $\lambda_2 = \frac{hc\lambda}{E\lambda + hc}$ .

Why other options are wrong:
- Option A: This option is incorrect due to an algebraic error in solving for $\lambda_2$ .
- Option C: This option is incorrect due to an algebraic error in solving for $\lambda_2$ .
- Option D: This option is incorrect due to an algebraic error in solving for $\lambda_2$ .

Key Concepts:
- Photoelectric effect
- Work function
- Kinetic energy
- Planck's constant

Memory Aids:
- Remember the photoelectric effect equation: $E = \frac{hc}{\lambda} - \phi$ .

Quick Tips:
- Always start with the photoelectric effect equation and then substitute the given values.

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The absolute zero is the temperature at which

View answer and explanation

Correct option: C

Molecular motion ceases

Option C (Molecular motion ceases) is correct because absolute zero (0 K or -273.15 °C) is theoretically the temperature at which all molecular motion stops.

Solution approach

Correct Answer + Core Reason: Molecular motion ceases. At absolute zero, molecular motion theoretically stops.

Step-by-Step Solution:

Step 1: Recall the definition of absolute zero.

Step 2: Absolute zero is the lowest possible temperature, defined as 0 Kelvin (0 K) or -273.15 degrees Celsius.

Step 3: At absolute zero, all molecular motion theoretically ceases, and the internal energy of a substance is at its minimum.

Why other options are wrong:
- Option A: Water freezes at 0 degrees Celsius, which is not absolute zero.
- Option B: Not all substances exist in solid state at absolute zero. Helium, for example, remains a liquid even at absolute zero under normal pressure.
- Option D: The correct answer is one of the options, so this is incorrect.

Key Concepts:
- Absolute Zero
- Temperature
- Molecular Motion
- Kinetic Energy

Memory Aids:
- Remember that absolute zero is the point of zero molecular motion.

Quick Tips:
- Absolute zero is a theoretical limit that is difficult to achieve in practice.

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The velocities of ten particles (in

\mathrm{ms}^{-1}

) are

0,2,3,4,4,4,5,5,6,9

. The most probable speed is

View answer and explanation

Correct option: B

4 \mathrm{~ms}^{-1}

Option B (4 ms⁻¹) is correct because the most probable speed is the speed that occurs most frequently in the given data set.

Solution approach

Correct Answer + Core Reason: 4 ms⁻¹. The most probable speed is the one that appears most often in the data set.

Step-by-Step Solution:

Step 1: Analyze the given velocities: 0, 2, 3, 4, 4, 4, 5, 5, 6, 9.

Step 2: Count the frequency of each velocity.

Step 3: Velocity 4 appears 3 times, which is more frequent than any other velocity.

Step 4: Therefore, the most probable speed is 4 ms⁻¹.

Why other options are wrong:
- Option A: 2 ms⁻¹ appears only once.
- Option C: 5 ms⁻¹ appears only twice.
- Option D: 9 ms⁻¹ appears only once.

Key Concepts:
- Most Probable Speed
- Data Analysis

Memory Aids:
- Most probable means most frequent.

Quick Tips:
- Simply count the occurrences of each value in the dataset.

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The kinetic energy per

g\;mol

for a diatomic gas at room temperature is

View answer and explanation

Correct option: B

\frac{5}{2}RT

Option B (

\frac{5}{2}RT

) is correct because a diatomic gas has 5 degrees of freedom at room temperature.

Solution approach

Correct Answer + Core Reason: $\frac{5}{2}RT$ is the kinetic energy per mole for a diatomic gas at room temperature.

Step-by-Step Solution:

Step 1: The kinetic energy per mole is given by $E = \frac{f}{2}RT$ , where $f$ is the number of degrees of freedom.

Step 2: For a diatomic gas at room temperature, the degrees of freedom are 3 translational and 2 rotational, so $f = 5$ .

Step 3: Therefore, $E = \frac{5}{2}RT$ .

Why other options are wrong:
- Option A: $3RT$ corresponds to 6 degrees of freedom, which is incorrect for a diatomic gas at room temperature.
- Option C: $\frac{3}{2}RT$ corresponds to 3 degrees of freedom, which is correct for a monoatomic gas.
- Option D: $\frac{1}{2}RT$ corresponds to 1 degree of freedom, which is incorrect for a diatomic gas.

Key Concepts:
- Kinetic Energy per Mole
- Degrees of Freedom
- Diatomic Gas

Memory Aids:
- Diatomic gas at room temperature: f = 5

Quick Tips:
- Remember the degrees of freedom for different types of gases.

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A vessel contains two non-reactive gases neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is

1: 2

. The ratio of number of molecules is:

View answer and explanation

Correct option: C

\frac{1}{2}

Option C (

\frac{1}{2}

) is correct because the ratio of the number of molecules is equal to the ratio of their partial pressures at the same volume and temperature.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}$ . The ratio of the number of molecules is equal to the ratio of partial pressures.

Step-by-Step Solution:

Step 1: Recall the ideal gas law: $PV = nRT$ , where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is temperature.

Step 2: For a mixture of non-reactive gases, the partial pressure of each gas is the pressure it would exert if it alone occupied the volume $V$ at temperature $T$ .

Step 3: Let $P_1$ be the partial pressure of neon and $P_2$ be the partial pressure of oxygen. Let $n_1$ be the number of moles of neon and $n_2$ be the number of moles of oxygen.

Step 4: Apply the ideal gas law to each gas: $P_1V = n_1RT$ and $P_2V = n_2RT$ .

Step 5: Divide the two equations: $\frac{{P_1V}}{{P_2V}} = \frac{{n_1RT}}{{n_2RT}} \Rightarrow \frac{{P_1}}{{P_2}} = \frac{{n_1}}{{n_2}}$ .

Step 6: Given that $\frac{{P_1}}{{P_2}} = \frac{1}{2}$ , we have $\frac{{n_1}}{{n_2}} = \frac{1}{2}$ .

Step 7: The number of molecules $N$ is related to the number of moles $n$ by Avogadro's number $N_A$ : $N = nN_A$ . Therefore, $\frac{{N_1}}{{N_2}} = \frac{{n_1N_A}}{{n_2N_A}} = \frac{{n_1}}{{n_2}} = \frac{1}{2}$ .

Why other options are wrong:
- **Option A ( $\frac{3}{2}$ ): This is incorrect and likely a result of a misunderstanding of the relationship between partial pressure and the number of molecules.
- Option B ( $\frac{2}{3}$ ): This is incorrect and likely a result of a misunderstanding of the relationship between partial pressure and the number of molecules.
- Option D ( $\frac{3}{5}$ ): This is incorrect and likely a result of a misunderstanding of the relationship between partial pressure and the number of molecules.

Key Concepts:
- Ideal Gas Law
- Partial Pressure
- Avogadro's Number

Memory Aids:
- Remember that at the same temperature and volume, the ratio of partial pressures is equal to the ratio of the number of moles (and therefore the number of molecules).

Quick Tips:**
- The ideal gas law is fundamental to understanding the behavior of gases.
- Partial pressures are additive: the total pressure is the sum of the partial pressures of all the gases in the mixture.

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The degrees of freedom of a triatomic gas is

View answer and explanation

Correct option: A

6

Option A (6) is correct because a triatomic gas molecule has 3 translational and 3 rotational degrees of freedom, totaling 6.

Solution approach

Correct Answer + Core Reason: 6. A triatomic gas has 3 translational and 3 rotational degrees of freedom.

Step-by-Step Solution:

Step 1: Define Degrees of Freedom. Degrees of freedom are the number of independent ways a molecule can store energy.

Step 2: Consider Translational Degrees of Freedom. A molecule can move in three independent directions (x, y, z), so it has 3 translational degrees of freedom.

Step 3: Consider Rotational Degrees of Freedom. A non-linear triatomic molecule can rotate about three independent axes, so it has 3 rotational degrees of freedom. A linear triatomic molecule has only 2 rotational degrees of freedom.

Step 4: Calculate Total Degrees of Freedom. For a non-linear triatomic molecule, the total degrees of freedom are 3 (translational) + 3 (rotational) = 6. For a linear triatomic molecule, it would be 3+2=5. Assuming the question refers to a non-linear triatomic molecule, the answer is 6.

Why other options are wrong:
- Option B: This is incorrect for a triatomic gas.
- Option C: This is incorrect for a triatomic gas.
- Option D: This is incorrect for a triatomic gas.

Key Concepts:
- Degrees of Freedom
- Translational Motion
- Rotational Motion
- Kinetic Theory of Gases

Memory Aids:
- Remember the degrees of freedom for different types of molecules.

Quick Tips:
- Distinguish between linear and non-linear molecules when calculating rotational degrees of freedom.

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Which one of the following is a wrong statement in kinetic theory of gases?

View answer and explanation

Correct option: B

The collision between molecules is inelastic.

Option B (The collision between molecules is inelastic) is correct because the kinetic theory of gases assumes perfectly elastic collisions between gas molecules.

Solution approach

Correct Answer + Core Reason: The collision between molecules is inelastic. The kinetic theory of gases assumes perfectly elastic collisions.

Step-by-Step Solution:

Step 1: State the postulates of the kinetic theory of gases.

Step 2: Identify the incorrect statement. The collision between molecules is assumed to be perfectly elastic, meaning kinetic energy is conserved during collisions.

Step 3: Explain why the other statements are correct. Gas molecules are in random motion, the volume occupied by the molecules is negligible compared to the volume of the container, and the force of attraction between molecules is negligible.

Why other options are wrong:
- Option A: Gas molecules are indeed in random motion.
- Option C: The volume occupied by the molecules is indeed negligible.
- Option D: The force of attraction between the molecules is indeed negligible.

Key Concepts:
- Kinetic theory of gases
- Elastic collisions
- Random motion

Memory Aids:
- Kinetic theory assumptions: Elastic collisions, negligible volume, negligible attraction

Quick Tips:
- Remember the key assumptions of the kinetic theory of gases.

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Which of the following is correct for the molecules of an ideal gas at a given temperature?

View answer and explanation

Correct option: B

Molecules have different speed distribution of which average remain constant

Option B (Molecules have different speed distribution of which average remain constant) is correct because, according to Maxwell's distribution, gas molecules at a given temperature have a range of speeds, but the average speed remains constant.

Solution approach

Correct Answer + Core Reason: Molecules have different speed distribution of which average remain constant. Maxwell's distribution describes a range of speeds with a constant average.

Step-by-Step Solution:

Step 1: Recall Maxwell's distribution of speeds.

Step 2: According to Maxwell's distribution, gas molecules at a given temperature have a range of speeds, not a single speed.

Step 3: The distribution is such that the average speed remains constant at a given temperature.

Step 4: Therefore, the correct statement is that molecules have different speed distributions, but the average remains constant.

Why other options are wrong:
- Option A: All have the same speed. This is incorrect due to Maxwell's distribution.
- Option C: Each molecule has a certain constant average speed. This is incorrect; each molecule's speed varies.
- Option D: They do not collide with one another. Ideal gas molecules do collide, although the collisions are perfectly elastic.

Key Concepts:
- Maxwell's distribution of speeds
- Kinetic theory of gases

Memory Aids:
- Maxwell's distribution shows a range of speeds.

Quick Tips:
- Remember that gas molecules have a distribution of speeds, not a single speed.

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At NTP, sample of equal volume of chlorine and oxygen is taken. Now ratio of no. of molecules is

View answer and explanation

Correct option: A

1\;:1

Option A (

1\;:\;1

) is correct because, according to Avogadro's hypothesis, equal volumes of gases at the same temperature and pressure contain the same number of molecules.

Solution approach

Correct Answer + Core Reason: $1\\;:\\;1$ . According to Avogadro's hypothesis, equal volumes of gases at the same temperature and pressure contain the same number of molecules.

Step-by-Step Solution:

Step 1: State Avogadro's hypothesis: Equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.

Step 2: The problem states that equal volumes of chlorine and oxygen are taken at NTP (Normal Temperature and Pressure).

Step 3: According to Avogadro's hypothesis, the number of molecules in equal volumes at the same temperature and pressure is the same.

Step 4: Therefore, the ratio of the number of molecules is 1:1.

Why other options are wrong:
- **Option B ( $32\\;:\\;27$ ): This ratio is incorrect and does not follow from Avogadro's hypothesis.
- Option C ( $2\\;:\\;1$ ): This ratio is incorrect and does not follow from Avogadro's hypothesis.
- Option D ( $16\\;:\\;14$ ): This ratio is incorrect and does not follow from Avogadro's hypothesis.

Key Concepts:
- Avogadro's hypothesis
- Molar volume of gases
- NTP conditions

Memory Aids:
- Remember Avogadro's number and its relation to the number of molecules in a mole of gas.

Quick Tips:**
- Apply Avogadro's hypothesis directly when dealing with equal volumes of gases at the same temperature and pressure.

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The number of degrees of freedom for each atom of a monoatomic gas is

View answer and explanation

Correct option: A

3

Option A (3) is correct because a monoatomic gas atom can move independently in three spatial dimensions (x, y, and z), giving it three translational degrees of freedom.

Solution approach

Correct Answer + Core Reason: 3. A monoatomic gas has 3 degrees of freedom due to translational motion.

Step-by-Step Solution:

Step 1: Define degrees of freedom.

Step 2: Degrees of freedom refer to the number of independent ways a molecule can store energy.

Step 3: For a monoatomic gas, the only way to store energy is through translational motion in three dimensions (x, y, and z).

Step 4: Therefore, a monoatomic gas has 3 degrees of freedom.

Why other options are wrong:
- Option B: 5 degrees of freedom is typical for diatomic molecules at higher temperatures (3 translational + 2 rotational).
- Option C: 6 degrees of freedom is not typical for monoatomic or diatomic gases at normal temperatures.
- Option D: 1 degree of freedom is not sufficient to describe the motion of a monoatomic gas in three-dimensional space.

Key Concepts:
- Degrees of freedom
- Monoatomic gas
- Translational motion

Memory Aids:
- Monoatomic = 3 (translational).

Quick Tips:
- Remember the degrees of freedom for different types of gases.

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A gas enclosed in a closed pot is kept in a train moving with high speed, the temperature of the gas

View answer and explanation

Correct option: C

Will remain the same

Option C is correct because the temperature of the gas depends on the average kinetic energy of its molecules, which is independent of the train's motion.

Solution approach

Correct Answer + Core Reason: Will remain the same. The temperature depends on the internal kinetic energy of the gas molecules, not the train's motion.

Step-by-Step Solution:

Step 1: Recall the definition of temperature: Temperature is a measure of the average kinetic energy of the molecules within a substance.

Step 2: Consider the frame of reference: The train's motion is an ordered motion of the entire system (pot and gas). The temperature depends on the disordered motion of the gas molecules relative to the pot.

Step 3: Since the train's motion does not affect the random motion of the gas molecules within the pot, the temperature remains the same.

Why other options are wrong:
- Option A: The train's motion does not directly increase the internal kinetic energy of the gas.
- Option B: The train's motion does not directly decrease the internal kinetic energy of the gas.
- Option D: The nature of the gas does not change due to the train's motion.

Key Concepts:
- Temperature
- Kinetic Energy
- Frame of Reference

Memory Aids:
- Temperature = Average Kinetic Energy of Molecules

Quick Tips:
- Distinguish between ordered and disordered motion.

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Assertion: Molecules of monoatomic gas has three degrees of freedom.

Reason: Molecules of diatomic gas has five degrees of freedom.

View answer and explanation

Correct option: B

If both assertion and reason are true and reason is not correct explanation of assertion.

Option B is correct because both statements are true, but the reason does not explain the assertion.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true and reason is not correct explanation of assertion. Both statements are correct, but not related.

Step-by-Step Solution:
Step 1: Analyze Assertion: A monoatomic gas molecule has 3 degrees of freedom (translational motion in x, y, and z directions).
Step 2: Analyze Reason: A diatomic gas molecule has 5 degrees of freedom (3 translational and 2 rotational).
Step 3: Conclude: Both assertion and reason are true, but the reason does not explain the assertion.

Why other options are wrong:
- Option A: The reason is not the correct explanation for the assertion.
- Option C: Both assertion and reason are true, so this option is incorrect.
- Option D: Both assertion and reason are true, so this option is incorrect.

Key Concepts:
- Degrees of Freedom
- Monoatomic Gas
- Diatomic Gas

Memory Aids:
- Monoatomic: 3 translational degrees of freedom.
- Diatomic: 3 translational + 2 rotational degrees of freedom.

Quick Tips:
- Remember the degrees of freedom for different types of gas molecules.
- Degrees of freedom relate to the ways a molecule can store energy.

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The temperature of an ideal gas is increased from

37^{\circ} \mathrm{C}

to

137^{\circ} \mathrm{C}

, then percentage increase in

v_\text{rms }

is:

View answer and explanation

Correct option: B

14.9 \%

Option B (

14.9 \%

) is correct because it is the calculated percentage increase in

v_\text{rms }

using the formula

v_{\text{rms}} = \sqrt{\frac{3RT}{M}}

and converting Celsius to Kelvin.

Solution approach

Correct Answer + Core Reason: $14.9 \%$ . The percentage increase in $v_\text{rms}$ is calculated using the formula $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ .

Step-by-Step Solution:

Step 1: Convert Celsius to Kelvin: $T_1 = 37^{\circ} \mathrm{C} + 273 = 310 \mathrm{K}$ and $T_2 = 137^{\circ} \mathrm{C} + 273 = 410 \mathrm{K}$ .

Step 2: Recall the formula for root-mean-square speed: $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ , where $R$ is the ideal gas constant and $M$ is the molar mass.

Step 3: Calculate the percentage increase in $v_{\text{rms}}$ :
$\text{Percentage increase} = \frac{v_{\text{rms2}} - v_{\text{rms1}}}{v_{\text{rms1}}} \times 100 = \frac{\sqrt{\frac{3RT_2}{M}} - \sqrt{\frac{3RT_1}{M}}}{\sqrt{\frac{3RT_1}{M}}} \times 100 = \frac{\sqrt{T_2} - \sqrt{T_1}}{\sqrt{T_1}} \times 100$ .

Step 4: Substitute the values of $T_1$ and $T_2$ :
$\text{Percentage increase} = \frac{\sqrt{410} - \sqrt{310}}{\sqrt{310}} \times 100 = \frac{20.248 - 17.607}{17.607} \times 100 = \frac{2.641}{17.607} \times 100 \approx 14.9 \%$ .

Step 5: State the final answer: The percentage increase in $v_{\text{rms}}$ is approximately $14.9 \%$ .

Why other options are wrong:
- **Option A ( $14 \%$ ): This is incorrect because it is a result of an approximate calculation or rounding errors.
- Option C ( $15.9 \%$ ): This is incorrect because it is a result of an approximate calculation or rounding errors.
- Option D ( $15 \%$ ): This is incorrect because it is a result of an approximate calculation or rounding errors.

Key Concepts:
- Root-mean-square speed
- Kinetic theory of gases
- Temperature conversion (Celsius to Kelvin)

Memory Aids:**
- Remember the formula for root-mean-square speed: $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$

Quick Tips:
- Always convert temperature to Kelvin when using gas laws.
- Pay attention to significant figures in calculations.

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What is the value of

\frac{R}{{{C_P}}}

for diatomic gas

View answer and explanation

Correct option: C

2/7

Option C is correct because for a diatomic gas, Cp = (7/2)R, so R/Cp = 2/7.

Solution approach

Correct Answer + Core Reason: $2/7$ . For a diatomic gas, $C_P = \frac{7}{2}R$ .

Step-by-Step Solution:

Step 1: Recall the value of $C_P$ for a diatomic gas: For a diatomic gas, the molar specific heat at constant pressure is $C_P = \frac{7}{2}R$ .

Step 2: Calculate the ratio $\frac{R}{C_P}$ : $\frac{R}{C_P} = \frac{R}{\frac{7}{2}R} = \frac{2}{7}$

Why other options are wrong:
- Option A: This is not the correct ratio for a diatomic gas.
- Option B: This is not the correct ratio for a diatomic gas.
- Option D: This is not the correct ratio for a diatomic gas.

Key Concepts:
- Molar specific heat
- Diatomic gas

Memory Aids:
- Remember that $C_P = \frac{7}{2}R$ for diatomic gases.

Quick Tips:
- Know the values of $C_P$ and $C_V$ for different types of gases.

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Three moles of nitrogen is mixed with six moles of helium. The effective molar specific heat of the mixture at constant volume is

View answer and explanation

Correct option: C

1.83 R

Option C (1.83 R) is correct because it accurately calculates the effective molar specific heat of the mixture using the formula

Cv_{mixture} = \frac{{n_1Cv_1 + n_2Cv_2}}{{n_1 + n_2}}

.

Solution approach

Correct Answer + Core Reason: 1.83 R. The effective molar specific heat is calculated using the weighted average of the individual molar specific heats.

Step-by-Step Solution:

Step 1: Identify the given information:
- Moles of nitrogen, $n_1 = 3$
- Moles of helium, $n_2 = 6$
- Molar specific heat of nitrogen (diatomic), $Cv_1 = \frac{{5R}}{2}$
- Molar specific heat of helium (monoatomic), $Cv_2 = \frac{{3R}}{2}$

Step 2: Calculate the effective molar specific heat of the mixture using the formula:
$Cv_{\text{{mixture}}} = \frac{{n_1Cv_1 + n_2Cv_2}}{{n_1 + n_2}}$

Step 3: Substitute the values:
$Cv_{\text{{mixture}}} = \frac{{3 \times \frac{{5R}}{2} + 6 \times \frac{{3R}}{2}}}{{3 + 6}}$

Step 4: Simplify the expression:
$Cv_{\text{{mixture}}} = \frac{{\frac{{15R}}{2} + \frac{{18R}}{2}}}{9} = \frac{{\frac{{33R}}{2}}}{9} = \frac{{33R}}{{18}}$

Step 5: Calculate the final value:
$Cv_{\text{{mixture}}} = \frac{{11R}}{6} \approx 1.83R$

Why other options are wrong:
- Option A: Incorrect calculation of the weighted average of molar specific heats.
- Option B: Incorrect calculation of the weighted average of molar specific heats.
- Option D: Incorrect calculation of the weighted average of molar specific heats.

Key Concepts:
- Molar specific heat at constant volume
- Mixture of gases
- Weighted average

Memory Aids:
- Remember the formula for the effective molar specific heat of a mixture.

Quick Tips:
- Identify the type of gas (monoatomic, diatomic) to determine the correct molar specific heat.
- Ensure correct substitution and simplification of the formula.

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The absolute temperature of a gas is determined by

View answer and explanation

Correct option: D

The mean square velocity of the molecules

Option D is correct because the absolute temperature of a gas is directly proportional to the mean square velocity of its molecules, as described by the kinetic theory of gases.

Solution approach

Correct Answer + Core Reason: The mean square velocity of the molecules. The absolute temperature is proportional to the mean square velocity of the molecules.

Step-by-Step Solution:

Step 1: Recall the kinetic theory of gases: The kinetic theory of gases relates the macroscopic properties of a gas (pressure, volume, temperature) to the microscopic properties of its molecules (mass, velocity).

Step 2: State the relationship between temperature and kinetic energy: The average translational kinetic energy of a gas molecule is directly proportional to the absolute temperature: $\frac{1}{2}m\overline{v^2} = \frac{3}{2}kT$ , where $m$ is the mass of the molecule, $\overline{v^2}$ is the mean square velocity, $k$ is the Boltzmann constant, and $T$ is the absolute temperature.

Step 3: Relate temperature to mean square velocity: From the equation above, we can see that $T = \frac{m}{3k}\overline{v^2}$ . Therefore, the absolute temperature is directly proportional to the mean square velocity of the molecules.

Step 4: Conclude that the absolute temperature of a gas is determined by the mean square velocity of the molecules.

Why other options are wrong:
- Option A: The average momentum is related to temperature, but the mean *square* velocity is the direct determinant.
- Option B: The velocity of sound depends on temperature, but temperature is determined by the mean square velocity.
- Option C: The number of molecules is related to the amount of gas, not directly to the temperature.

Key Concepts:
- Kinetic theory of gases
- Absolute temperature
- Mean square velocity
- Boltzmann constant

Memory Aids:
- Temperature is a measure of average kinetic energy.

Quick Tips:
- Remember that the absolute temperature is proportional to the mean square velocity of the molecules.

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