KINETIC THEORY OF GASES MCQs for JEE PHYSICS — Medium, Page 7

181 Medium

181 of 257

Understanding +4 | -1

The ratio of specific heats of an ideal gas is

View answer and explanation

Correct option: B

1+\frac{R}{C_v}

Option B (

1+\frac{R}{C_v}

) is correct because it represents the ratio of specific heats (

\gamma = \frac{C_P}{C_V}

) derived from the relationship

C_P - C_V = R

.

Solution approach

Correct Answer + Core Reason: $1+\frac{R}{C_v}$ . This formula is derived from the relation between $C_P$ , $C_V$ , and $R$ .

Step-by-Step Solution:

Step 1: Recall the relationship between specific heats at constant pressure ( $C_P$ ) and constant volume ( $C_V$ ) and the gas constant ( $R$ ). The relationship is given by $C_P - C_V = R$ .

Step 2: Define the ratio of specific heats ( $\gamma$ ). The ratio of specific heats is defined as $\gamma = \frac{{C_P}}{{C_V}}$ .

Step 3: Rearrange the equation $C_P - C_V = R$ to express $C_P$ in terms of $C_V$ and $R$ . We have $C_P = C_V + R$ .

Step 4: Substitute this expression for $C_P$ into the definition of $\gamma$ . We get $\gamma = \frac{{C_V + R}}{{C_V}} = 1 + \frac{R}{{C_V}}$ .

Why other options are wrong:
- Option A: $\frac{1}{1-\frac{R}{C_{\mathrm{P}}}}$ . This is incorrect. It would be correct if we were using $C_P$ instead of $C_V$ in the final expression.
- Option C: $\frac{1}{1-\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{R}}}$ . This is incorrect. It's a misapplication of the relationship between $C_P$ , $C_V$ and $R$ .
- Option D: $\frac{C_{\mathrm{V}}}{C_{\mathrm{P}}}+\mathrm{R}$ . This is incorrect. It does not represent the ratio of specific heats.

Key Concepts:
- Specific heat at constant pressure ( $C_P$ )
- Specific heat at constant volume ( $C_V$ )
- Gas constant ( $R$ )
- Ratio of specific heats ( $\gamma$ )

Memory Aids:
- Remember the relationship: $C_P - C_V = R$

Quick Tips:
- Always remember the relationship between $C_P$ , $C_V$ , and $R$ .

182 Medium

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Application +4 | -1

The kinetic energy of one g-mole of a gas at normal temperature and pressure is (

R = 8.31\;J/mol - K

)

View answer and explanation

Correct option: D

3.4 \times {10^3}J

Option D (

3.4 \times {10^3}J

) is correct because it is the kinetic energy calculated using the formula

E = \frac{3}{2}RT

with the given values.

Solution approach

Correct Answer + Core Reason: $3.4 \times {10^3}J$ . This is the kinetic energy calculated using the formula $E = \frac{3}{2}RT$ .

Step-by-Step Solution:

Step 1: Identify the given parameters: $R = 8.31 \text{{ J/mol K}}$ , $T = 273 \text{{ K}}$ (Normal Temperature).

Step 2: Use the formula for the kinetic energy of one mole of a gas: $E = \frac{3}{2}RT$ .

Step 3: Substitute the values into the formula: $E = \frac{3}{2} \times 8.31 \times 273$ .

Step 4: Calculate the kinetic energy: $E = 1.5 \times 8.31 \times 273 = 12.465 \times 273 = 3403.945 \approx 3.4 \times 10^3 \text{{ J}}$ .

Why other options are wrong:
- Option A: $0.56 \times {10^4}J$ - This option is incorrect because it overestimates the kinetic energy.
- Option B: $1.3 \times {10^2}J$ - This option is incorrect because it significantly underestimates the kinetic energy.
- Option C: $2.7 \times {10^2}J$ - This option is incorrect because it underestimates the kinetic energy.

Key Concepts:
- Kinetic energy of a gas
- Ideal gas constant
- Temperature

Memory Aids:
- Remember the formula for kinetic energy of one mole of a gas: $E = \frac{3}{2}RT$ .

Quick Tips:
- Make sure to use the correct units for all parameters.
- Remember the value of the ideal gas constant $R$ .

183 Medium

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Application +4 | -1

What is the ratio of rms speed of oxygen and hydrogen molecules at the same temperature?

View answer and explanation

Correct option: C

1:4

The correct answer is C (1:4) because the rms speed is inversely proportional to the square root of the molar mass.

Solution approach

Correct Answer + Core Reason: 1:4. The ratio of rms speeds is inversely proportional to the square root of the molar masses.

Step-by-Step Solution:

Step 1: ${V_{{\text{rms}}}} = \sqrt {\frac{{3RT}}{M}} \Rightarrow {V_{{\text{rms}}}} \propto \frac{1}{{\sqrt M }}$

Step 2: $\frac{{{V_{{{{O_2}}}}}}}{{{V_{{{{H_2}}}}}}} = \sqrt {\frac{{{M_{{{{H_2}}}}}}}{{{M_{{{{O_2}}}}}}}} = \sqrt {\frac{2}{{32}}} = \frac{1}{4}$

Why other options are wrong:
- Option A: Incorrect calculation.
- Option B: Incorrect calculation.
- Option D: Incorrect calculation, inverted the ratio.

Key Concepts:
- Root mean square speed
- Kinetic theory of gases

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

Quick Tips:
- Pay attention to the molar masses of the gases.

184 Medium

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Understanding +4 | -1

At constant pressure, which of the following is true?

View answer and explanation

Correct option: D

v \propto \frac{1}{{\sqrt \rho }}

Option D (

v \propto \frac{1}{{\sqrt \rho }}

) is correct because at constant pressure, the speed of sound is inversely proportional to the square root of the density.

Solution approach

Correct Answer + Core Reason: $v \propto \frac{1}{{\sqrt \rho }}$ . At constant pressure, the speed of sound is inversely proportional to the square root of the density.

Step-by-Step Solution:

Step 1: Recall the formula for pressure due to an ideal gas: $p = \frac{M}{{3V}}{v^2}$

Step 2: Express density as $\rho = \frac{M}{V}$

Step 3: Substitute density into the pressure formula: $p = \frac{1}{3}\rho {v^2}$

Step 4: Solve for $v$ : $v = \sqrt {\left( {\frac{{3p}}{\rho }} \right)}$

Step 5: At constant pressure, $v \propto \frac{1}{{\sqrt \rho }}$

Why other options are wrong:
- Option A: $v \propto \sqrt \rho$ is incorrect because the speed of sound is inversely proportional to the square root of the density.
- Option B: $v \propto \frac{1}{\rho }$ is incorrect because the speed of sound is inversely proportional to the square root of the density, not the density itself.
- Option C: $v \propto \rho$ is incorrect because the speed of sound is inversely proportional to the square root of the density.

Key Concepts:
- Speed of sound in a gas
- Ideal gas law
- Density

Memory Aids:
- Remember the relationship between speed of sound and density at constant pressure.

Quick Tips:
- The speed of sound depends on the properties of the medium.

185 Medium

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Application +4 | -1

The speeds of 5 molecules of a gas (in arbitrary units) are as follows:

2,3,4,5,6.

The root mean square speed for these molecules is

View answer and explanation

Correct option: D

4.24

Option D (4.24) is correct because it is the root mean square speed calculated using the given speeds of the molecules.

Solution approach

Correct Answer + Core Reason: 4.24. The root mean square speed is calculated using the formula $v_{rms} = \sqrt{\frac{\sum v_i^2}{N}}$ .

Step-by-Step Solution:

Step 1: The speeds of the 5 molecules are given as 2, 3, 4, 5, 6.

Step 2: The root mean square speed is calculated as: $v_{rms} = \sqrt{\frac{2^2 + 3^2 + 4^2 + 5^2 + 6^2}{5}}$ .

Step 3: Calculating the sum of the squares: $2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 4 + 9 + 16 + 25 + 36 = 90$ .

Step 4: Dividing by the number of molecules: $\frac{90}{5} = 18$ .

Step 5: Taking the square root: $v_{rms} = \sqrt{18} = 4.24$ .

Why other options are wrong:
- Option A: 2.91 is incorrect because it is not the root mean square speed. It might be a simple average or some other incorrect calculation.
- Option B: 3.52 is incorrect because it is not the root mean square speed. It might be a weighted average or some other incorrect calculation.
- Option C: 4 is incorrect because it is close to the average but not the root mean square speed.

Key Concepts:
- Root Mean Square Speed: $v_{rms} = \sqrt{\frac{\sum v_i^2}{N}}$

Memory Aids:
- Remember the formula for root mean square speed.

Quick Tips:
- Square each speed before summing them up.
- Divide by the number of molecules before taking the square root.

186 Medium

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Understanding +4 | -1

Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafluoride (polyatomic). Arrange these on the basis of their root mean square speed

\left(v_{\mathrm{ms}}\right)

and choose the correct answer from the options given below:

View answer and explanation

Correct option: A

v_{\text {rms }}

(mono)

>v_{\mathrm{ms}}

(dia)

>v_{\mathrm{rms}}

(poly)

Option A (

v_{\text {rms }}

(mono)

>v_{\mathrm{ms}}

(dia)

>v_{\mathrm{rms}}

(poly)) is correct because the root mean square speed is inversely proportional to the square root of the molar mass.

Solution approach

Correct Answer + Core Reason: $v_{\text {rms }}$ (mono) $>v_{\mathrm{ms}}$ (dia) $>v_{\mathrm{rms}}$ (poly). RMS speed is inversely proportional to the square root of molar mass.

Step-by-Step Solution:

Step 1: The root mean square speed is given by the formula: $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ , where $R$ is the gas constant, $T$ is the temperature, and $M$ is the molar mass.

Step 2: Since the temperature is the same for all three gases, $v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$ .

Step 3: The molar masses are: Neon (monoatomic) < Chlorine (diatomic) < Uranium hexafluoride (polyatomic).

Step 4: Therefore, $v_{\text{rms}}$ (mono) $> v_{\text{rms}}$ (dia) $> v_{\text{rms}}$ (poly).

Why other options are wrong:
- Option B: $v_{\mathrm{rms}}($ mono $)=\mathrm{v}_{\mathrm{ms}}($ dia $)=\mathrm{v}_{\mathrm{ms}}($ poly): This is incorrect because the molar masses are different.
- Option C: $v_{\text {rms }}$ (mono) $< v_{\mathrm{ms}}$ (dia) $< v_{\text {rms }}$ (poly): This is incorrect because the root mean square speed is inversely proportional to the square root of the molar mass.
- Option D: $v_{\text {rms }}$ (dia) $< v_{\text {rms }}$ (poly) $< v_{\text {rms }}$ (mono): This is incorrect because the root mean square speed is inversely proportional to the square root of the molar mass.

Key Concepts:
- Root mean square speed
- Molar mass
- Kinetic theory of gases

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

Quick Tips:
- Consider the molar mass of the gas when comparing root mean square speeds.

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Analyzing +4 | -1

Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is

1: 4

, then
A. The r.m.s. velocity of gas molecules in two vessels will be the same.
B. The ratio of pressure in these vessels will be

1 : 4

.
C. The ratio of pressure will be

1 : 1

.
D. The r.m.s. velocity of gas molecules in two vessels will be in the ratio of

1 : 4

.
Choose the correct answer from the options given
below

View answer and explanation

Correct option: C

A and B only

Option C (A and B only) is correct because the r.m.s. velocity is the same due to the same temperature, and the pressure ratio is 1:4 due to the number of molecules ratio.

Solution approach

Correct Answer + Core Reason: A and B only. The r.m.s. velocity depends on temperature, and pressure is proportional to the number of molecules.

Step-by-Step Solution:

Step 1: Identify the given information: Same gas, same volume, same temperature, and the ratio of the number of molecules is 1:4.

Step 2: Recall the formula for r.m.s. velocity: $v_{\mathrm{rms}} = \sqrt{\frac{{3RT}}{{M_0}}}$

Step 3: Since the gas and temperature are the same, the r.m.s. velocity will be the same.

Step 4: Recall the ideal gas law: $PV = nRT$ , where $n$ is the number of moles, which is proportional to the number of molecules.

Step 5: Calculate the ratio of pressures: $\frac{{P_1}}{{P_2}} = \frac{{n_1RT/V}}{{n_2RT/V}} = \frac{{n_1}}{{n_2}} = \frac{1}{4}$

Step 6: Conclude that the r.m.s. velocity is the same, and the ratio of pressures is 1:4.

Why other options are wrong:
- Option A: Incorrect because it includes C, which states the ratio of pressure will be 1:1, which is wrong.
- Option B: Incorrect because it includes D, which states the r.m.s. velocity will be in the ratio of 1:4, which is wrong.
- Option D: Incorrect because it includes C and D, both of which are wrong.

Key Concepts:
- Root Mean Square (RMS) velocity
- Ideal Gas Law

Memory Aids:
- RMS velocity depends only on temperature for a given gas.
- Ideal gas law: PV = nRT

Quick Tips:
- Remember that r.m.s. velocity is directly proportional to the square root of temperature.
- Pressure is directly proportional to the number of molecules at constant volume and temperature.

188 Medium

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Application +4 | -1

A vessel contains

1.60 \mathrm{~g}

of oxygen and

2.80 \mathrm{~g}

of nitrogen. The temperature is maintained at

300 \mathrm{~K}

and the volume of the vessel is

0.166 \mathrm{~m}^3

. Find the pressure of the mixture.

View answer and explanation

Correct option: B

2250 \mathrm{~N/m^2}

The correct answer is 2250 N/m² because the pressure of the mixture is calculated using the ideal gas law and Dalton's law of partial pressures.

Solution approach

Correct Answer + Core Reason: $2250 \text{{ N/m}}^2$ . The pressure is calculated using the ideal gas law and Dalton's law of partial pressures.

Step-by-Step Solution:

Step 1: Identify the given information: Mass of oxygen $m_{O_2} = 1.60 \text{{ g}}$ , Mass of nitrogen $m_{N_2} = 2.80 \text{{ g}}$ , Temperature $T = 300 \text{{ K}}$ , Volume $V = 0.166 \text{{ m}}^3$ .

Step 2: Calculate the number of moles of oxygen: $n_{O_2} = \frac{m_{O_2}}{M_{O_2}} = \frac{1.60 \text{{ g}}}{32 \text{{ g/mol}}} = 0.05 \text{{ mol}}$ .

Step 3: Calculate the number of moles of nitrogen: $n_{N_2} = \frac{m_{N_2}}{M_{N_2}} = \frac{2.80 \text{{ g}}}{28 \text{{ g/mol}}} = 0.1 \text{{ mol}}$ .

Step 4: Calculate the total number of moles: $n = n_{O_2} + n_{N_2} = 0.05 + 0.1 = 0.15 \text{{ mol}}$ .

Step 5: Apply the ideal gas law: $PV = nRT$ , where $R = 8.3 \text{{ J/(mol K)}}$ .

Step 6: Solve for the pressure: $P = \frac{nRT}{V} = \frac{0.15 \text{{ mol}} \times 8.3 \text{{ J/(mol K)}} \times 300 \text{{ K}}}{0.166 \text{{ m}}^3} = \frac{373.5}{0.166} \approx 2250 \text{{ N/m}}^2$ .

Final Answer: $2250 \text{{ N/m}}^2$

Why other options are wrong:
- Option A (2350 N/m²): This is incorrect. It might be calculated with slight rounding errors or incorrect substitution.
- Option C (2450 N/m²): This is incorrect. It might be calculated with significant rounding errors or incorrect substitution.
- Option D (2550 N/m²): This is incorrect. It might be calculated with significant rounding errors or incorrect substitution.

Key Concepts:
- Ideal gas law: $PV = nRT$
- Dalton's law of partial pressures
- Moles calculation

Memory Aids:
- PV = nRT (Ideal Gas Law)

Quick Tips:
- Make sure to use consistent units (SI units).
- Remember to calculate the total number of moles for the mixture.

189 Medium

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Application +4 | -1

Assume that the temperature remains essentially constant in the upper part of the atmosphere. Obtain an expression for the variation in pressure in the upper atmosphere with height. The mean molecular weight of air is

M

.

View answer and explanation

Correct option: D

p=p_0 e^{-\frac{M g h}{R T}}

Option D (

p=p_0 e^{-\frac{M g h}{R T}}

) is correct because it accurately describes the exponential decrease of pressure with height in the upper atmosphere, considering constant temperature.

Solution approach

Correct Answer + Core Reason: $p=p_0 e^{-\frac{M g h}{R T}}$ . This equation describes the pressure variation with height in an isothermal atmosphere.

Step-by-Step Solution:

Step 1: Consider the pressure difference $dp$ over a small height $dh$ : $dp = -\rho g dh$ , where $\rho$ is the density of air.

Step 2: Use the ideal gas law to relate pressure and density: $pV = nRT = \frac{m}{M}RT$ , where $M$ is the molar mass. Therefore, $p = \frac{\rho}{M}RT$ , and $\rho = \frac{Mp}{RT}$ .

Step 3: Substitute the expression for $\rho$ into the pressure difference equation: $dp = -\frac{Mg}{RT}p dh$ .

Step 4: Separate variables and integrate: $\int_{p_0}^{p} \frac{dp}{p} = -\int_{0}^{h} \frac{Mg}{RT} dh$ , where $p_0$ is the pressure at $h=0$ .

Step 5: Evaluate the integrals: $\ln{\frac{p}{p_0}} = -\frac{Mgh}{RT}$ .

Step 6: Solve for $p$ : $p = p_0 e^{-\frac{Mgh}{RT}}$ .

Why other options are wrong:
- **Option A ( $p=p_0 e^{\frac{M g h}{R T}}$ ): This is incorrect because it shows pressure increasing with height, which is not physically accurate.
- Option B ( $p=p_0 e^{-\frac{RT}{Mgh}}$ ): This is incorrect because the exponent is inverted, and the equation does not correctly represent the relationship between pressure and height.
- Option C ( $p=p_0 e^{\frac{RT}{Mgh}}$ ): This is incorrect because the exponent is inverted, and it shows pressure increasing with height, which is not physically accurate.

Key Concepts:
- Ideal gas law
- Hydrostatic equilibrium
- Exponential decay

Memory Aids:
- Remember that pressure decreases with height in the atmosphere.

Quick Tips:**
- Always consider the sign when dealing with pressure gradients.
- Use the ideal gas law to relate pressure, density, and temperature.

190 Medium

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Application +4 | -1

At room temperature (27

^ \circ C)\;

the rms speed of the molecules of a certain diatomic gas is found to be 1920 m

{s^{ - 1}}

. The gas is

View answer and explanation

Correct option: D

{\mathrm{H}_2}

Option D (

{\mathrm{H}_2}

) is correct because the calculated molar mass from the given rms speed matches that of hydrogen gas.

Solution approach

Correct Answer + Core Reason: ${\mathrm{H}_2}$ . The molar mass calculated from the rms speed matches hydrogen.

Step-by-Step Solution:

Step 1: Identify the given information: rms speed $C_{\mathrm{rms}} = 1920 \text{{ m/s}}$ , temperature $T = 27^\circ C = 300 \text{{ K}}$ .

Step 2: Recall the formula for rms speed: $C_{\mathrm{rms}} = \sqrt{\frac{{3RT}}{M}}$ , where $R$ is the ideal gas constant and $M$ is the molar mass.

Step 3: Rearrange the formula to solve for molar mass: $M = \frac{{3RT}}{{C_{\mathrm{rms}}^2}}$

Step 4: Substitute the given values: $M = \frac{{3 \times 8.31 \times 300}}{{{{\left( {1920} \right)}^2}}}$

Step 5: Calculate the molar mass: $M = \frac{{7479}}{{3686400}} = 0.00203 \text{{ kg/mol}} = 2.03 \text{{ g/mol}}$

Step 6: Compare the calculated molar mass to the molar masses of the given gases.

Step 7: The molar mass of hydrogen ( ${\mathrm{H}_2}$ ) is approximately 2 g/mol, which matches the calculated value.

Why other options are wrong:
- **Option A ( ${\mathrm{C}{\mathrm{l}_2}}$ ): Chlorine has a molar mass of approximately 71 g/mol, which is significantly different from the calculated value.
- Option B ( ${\mathrm{O}_2}$ ): Oxygen has a molar mass of approximately 32 g/mol, which is significantly different from the calculated value.
- Option C ( ${\mathrm{N}_2}$ ): Nitrogen has a molar mass of approximately 28 g/mol, which is significantly different from the calculated value.

Key Concepts:
- Root Mean Square (RMS) speed of gas molecules
- Ideal gas constant
- Molar mass

Memory Aids:**
- Remember the formula: $C_{\mathrm{rms}} = \sqrt{\frac{{3RT}}{M}}$

Quick Tips:
- Ensure consistent units (SI units) before calculations.
- Remember the value of the ideal gas constant, $R = 8.31 \text{{ J/(mol K)}}$ .
- Convert temperature to Kelvin.

191 Medium

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Understanding +4 | -1

The ratio of

\mathrm{C}_{\mathrm{p}}

and

\mathrm{C}_{\mathrm{v}}

depends on temperature

\mathrm{T}

according to relation

View answer and explanation

Correct option: A

\gamma \propto \mathrm{T}

The correct answer is that

\gamma

does not depend on temperature, so none of the options are correct. The provided answer and explanation are incorrect.

Solution approach

Correct Answer + Core Reason: None of the options are correct. The ratio of specific heats, $\gamma$ , is generally independent of temperature for ideal gases.

Step-by-Step Solution:

Step 1: Define the ratio of specific heats: $\gamma = \frac{C_p}{C_v}$ , where $C_p$ is the specific heat at constant pressure and $C_v$ is the specific heat at constant volume.

Step 2: Consider ideal gases: For ideal gases, $C_p$ and $C_v$ are related by $C_p = C_v + R$ , where R is the gas constant. Both $C_p$ and $C_v$ can depend on temperature, but their ratio is often approximately constant, especially for monatomic and diatomic gases over a wide range of temperatures.

Step 3: Analyze the temperature dependence: While $C_p$ and $C_v$ can have some temperature dependence, the ratio $\gamma$ is often treated as a constant in many thermodynamic calculations, particularly for ideal gases. For real gases at high temperatures, $\gamma$ can exhibit some temperature dependence due to vibrational modes becoming significant.

Why other options are wrong:
- Option A: Incorrect. $\gamma$ is not directly proportional to T.
- Option B: Incorrect. $\gamma$ is not equal to T.
- Option C: Incorrect. $\gamma$ is not inversely proportional to T.
- Option D: Incorrect. $\gamma$ is not proportional to the square root of T.

Key Concepts:
- Specific heat at constant pressure ( $C_p$ )
- Specific heat at constant volume ( $C_v$ )
- Ratio of specific heats ( $\gamma$ )
- Ideal gas law
- Temperature dependence of specific heats

Memory Aids:
- $\gamma = \frac{C_p}{C_v}$ (Remember the definition)

Quick Tips:
- For ideal gases, assume $\gamma$ is constant unless otherwise specified.
- Be aware that for real gases at high temperatures, $\gamma$ can have some temperature dependence.

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Understanding +4 | -1

Mean free path is inversely proportional to

View answer and explanation

Correct option: D

Both 1 and 2

Option D (Both 1 and 2) is correct because the mean free path is inversely proportional to both the number density and the square of the molecular diameter.

Solution approach

Correct Answer + Core Reason: Both 1 and 2. Mean free path is inversely proportional to both number density and the square of molecular diameter.

Step-by-Step Solution:

Step 1: Recall the formula for mean free path: $\lambda = \frac{1}{{\sqrt{2} \pi d^2 n}}$ , where $d$ is the molecular diameter and $n$ is the number density.

Step 2: From the formula, it is clear that $\lambda \propto \frac{1}{n}$ and $\lambda \propto \frac{1}{d^2}$ .

Step 3: Therefore, the mean free path is inversely proportional to both the number density and the square of the molecular diameter.

Step 4: Final Answer: Both 1 and 2

Why other options are wrong:
- Option A: Number density: This is only partially correct, as the square of the molecular diameter also plays a role.
- Option B: Square of molecular diameter: This is only partially correct, as the number density also plays a role.
- Option C: Square of number of density: This is incorrect because the mean free path is inversely proportional to the number density, not its square.

Key Concepts:
- Mean Free Path
- Number Density
- Molecular Diameter

Memory Aids:
- Remember the formula for mean free path.

Quick Tips:
- Understand the relationship between mean free path, number density, and molecular diameter.

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Understanding +4 | -1

If the rms velocity of a gas is

v

,then

View answer and explanation

Correct option: B

{v^2}/T = \mathrm{constant}

Option B (

{v^2}/T = \mathrm{constant}

) is correct because the square of the RMS velocity is directly proportional to the temperature.

Solution approach

Correct Answer + Core Reason: ${v^2}/T = \mathrm{constant}$ . The RMS velocity squared is proportional to temperature.

Step-by-Step Solution:

Step 1: The root-mean-square (RMS) velocity of a gas is given by the formula: $v = \sqrt{{\frac{{3kT}}{m}}}$ , where k is Boltzmann's constant, T is the absolute temperature, and m is the mass of a molecule.

Step 2: Squaring both sides of the equation, we get: ${v^2} = \frac{{3kT}}{m}$ .

Step 3: Rearranging the equation, we have: $\frac{{v^2}}{T} = \frac{{3k}}{m}$ .

Step 4: Since k and m are constants for a given gas, $\frac{{3k}}{m}$ is also a constant. Therefore, $\frac{{v^2}}{T} = \mathrm{constant}$ .

Why other options are wrong:
- Option A: This is incorrect because it states that ${v^2}T$ is constant, which is the inverse of the correct relationship.
- Option C: This is incorrect because it states that $v{T^2}$ is constant, which is not the correct relationship.
- Option D: This is incorrect because the RMS velocity is dependent on temperature.

Key Concepts:
- Root-Mean-Square (RMS) Velocity
- Kinetic Theory of Gases
- Boltzmann's Constant

Memory Aids:
- Remember the formula for RMS velocity: $v = \sqrt{{\frac{{3kT}}{m}}}$ .

Quick Tips:
- Always remember that temperature must be in Kelvin when using the RMS velocity formula.

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Application +4 | -1

If mass of gas is double and

v_{r m s}

is half then pressure of gas will be.

View answer and explanation

Correct option: A

P^{\prime} \propto \frac{1}{2} \text { times }

Option A (

P^{\prime} \propto \frac{1}{2} \text { times }

) is correct because pressure is proportional to mass and the square of the RMS velocity, and the changes in these variables result in a halving of the pressure.

Solution approach

Correct Answer + Core Reason: $P^{\prime} \propto \frac{1}{2} \text { times }$ . Pressure is proportional to mass and the square of RMS velocity.

Step-by-Step Solution:

Step 1: Write the formula for pressure in terms of density and RMS velocity: $P = \frac{1}{3} \rho v_{rms}^2$

Step 2: Express density in terms of mass and volume: $\rho = \frac{M}{V}$

Step 3: Substitute the expression for density into the pressure formula: $P = \frac{1}{3} \frac{M}{V} v_{rms}^2$

Step 4: Identify the proportionality: $P \propto M v_{rms}^2$

Step 5: Define the new pressure $P^{\prime}$ in terms of the new mass $M^{\prime} = 2M$ and new RMS velocity $v_{rms}^{\prime} = \frac{1}{2} v_{rms}$ : $P^{\prime} \propto M^{\prime} (v_{rms}^{\prime})^2 = 2M (\frac{1}{2} v_{rms})^2$

Step 6: Simplify the expression for $P^{\prime}$ : $P^{\prime} \propto 2M \frac{1}{4} v_{rms}^2 = \frac{1}{2} M v_{rms}^2$

Step 7: Relate the new pressure to the original pressure: $P^{\prime} \propto \frac{1}{2} P$

Final Answer: $P^{\prime} \propto \frac{1}{2} \text { times }$

Why other options are wrong:
- Option B: This is incorrect because it calculates the pressure as 3/2 times the original, which is not supported by the change in mass and RMS velocity.
- Option C: This is incorrect because it calculates the pressure as 3/4 times the original, which is not supported by the change in mass and RMS velocity.
- Option D: This is incorrect because it calculates the pressure as 5/4 times the original, which is not supported by the change in mass and RMS velocity.

Key Concepts:
- Pressure of a gas
- Root mean square velocity
- Density
- Proportionality

Memory Aids:
- Remember the formula: $P = \frac{1}{3} \rho v_{rms}^2$

Quick Tips:
- Pay attention to how each variable affects the pressure.
- When dealing with proportionality, express the new variables in terms of the original variables.

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Understanding +4 | -1

At

{0^0}C

temperature, root mean square speed of which of the following gas will be maximum?

View answer and explanation

Correct option: A

{H_2}

Option A (

{H_2}

) is correct because the root mean square speed is inversely proportional to the square root of the molar mass, and hydrogen has the lowest molar mass.

Solution approach

Correct Answer + Core Reason: ${H_2}$ . Root mean square speed is inversely proportional to the square root of the molar mass.

Step-by-Step Solution:

Step 1: Identify the formula for root mean square speed: ${V_{rms}} = \sqrt{\frac{{3RT}}{M}}$ where R is the gas constant, T is the temperature, and M is the molar mass.

Step 2: Analyze the formula: At a constant temperature, ${V_{rms}} \propto \frac{1}{{\sqrt{M}}}$ . This means the gas with the smallest molar mass will have the highest root mean square speed.

Step 3: Compare the molar masses of the given gases: ${H_2}$ (2 g/mol), ${N_2}$ (28 g/mol), ${O_2}$ (32 g/mol).

Step 4: Determine the gas with the smallest molar mass: ${H_2}$ has the smallest molar mass.

Step 5: Conclusion: Therefore, ${H_2}$ will have the maximum root mean square speed.

Why other options are wrong:
- **Option B ( ${N_2}$ ): Nitrogen has a higher molar mass than hydrogen, so its root mean square speed will be lower.
- Option C ( ${O_2}$ ): Oxygen has a higher molar mass than hydrogen, so its root mean square speed will be lower.
- Option D (Same for all gases): This is incorrect because the root mean square speed depends on the molar mass of the gas.

Key Concepts:
- Root mean square speed
- Kinetic theory of gases
- Molar mass

Memory Aids:**
- Remember the formula: ${V_{rms}} = \sqrt{\frac{{3RT}}{M}}$

Quick Tips:
- The lighter the gas, the faster its molecules move at a given temperature.

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Understanding +4 | -1

Assertion: For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gas in the mixture.
Reason: In equilibrium, the average kinetic energy of the molecules of different gases will be equal.

View answer and explanation

Correct option: A

If both assertion and reason are true and reason is the correct explanation of assertion

Option A (If both assertion and reason are true and reason is the correct explanation of assertion) is correct because Dalton's law is based on the equal average kinetic energy of gas molecules at equilibrium.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true and reason is the correct explanation of assertion. Dalton's Law is a direct consequence of the kinetic theory of gases.

Step-by-Step Solution:

Step 1: Understand Dalton's Law: The total pressure of a mixture of non-reactive ideal gases is the sum of the partial pressures of each gas.

Step 2: Understand the Kinetic Theory of Gases: At a given temperature, the average kinetic energy of gas molecules is the same, regardless of the gas type.

Step 3: Connect the concepts: The partial pressure of each gas is proportional to its number density and the average kinetic energy of its molecules. Since the average kinetic energy is the same for all gases at equilibrium, the total pressure is simply the sum of the partial pressures.

Why other options are wrong:
- Option B: While both statements are true, the reason *is* the correct explanation of the assertion.
- Option C: The reason is true, so this option is incorrect.
- Option D: Both the assertion and the reason are true, so this option is incorrect.

Key Concepts:
- Dalton's Law of Partial Pressures
- Kinetic Theory of Gases
- Average Kinetic Energy

Memory Aids:
- Dalton's Law: Total pressure = Sum of partial pressures
- Kinetic Theory: Same temperature = Same average KE

Quick Tips:
- Remember that Dalton's Law applies to ideal gases that do not react chemically.

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Understanding +4 | -1

Critical temperature of

\mathrm{CO}_2

is

31.2^{\circ} \mathrm{C}

. In summer, the room temperature is

40^{\circ} \mathrm{C}

.

View answer and explanation

Correct option: A

\mathrm{CO}_2

cannot be liquefied

Option A is correct because the pressure of an ideal gas is independent of the container's shape and depends on the number of moles, temperature, and volume, as described by the ideal gas law.

Solution approach

Correct Answer + Core Reason: If both Assertion and reason are true and the Reason is the correct explanation of the Assertion. The ideal gas law explains why pressure is independent of shape and dependent on moles, temperature, and volume.

Step-by-Step Solution:

Step 1: State 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: Analyze the Ideal Gas Law: From the equation, we can see that pressure $P$ depends on $n$ , $R$ , $T$ , and $V$ . It does not depend on the shape of the container.

Step 3: Explain the Assertion: The assertion states that the pressure exerted by an enclosed ideal gas does not depend on the shape of the container. This is true because the shape of the container does not appear in the ideal gas law.

Step 4: Explain the Reason: The reason states that the pressure of an ideal gas depends on the number of moles, temperature, and volume of the enclosed gas. This is also true and is directly supported by the ideal gas law.

Step 5: Conclusion: The reason correctly explains the assertion because the ideal gas law shows that pressure depends on $n$ , $T$ , and $V$ , not on the shape of the container.

Why other options are wrong:
- Option B: The reason *is* the correct explanation of the assertion.
- Option C: The reason is not false; it accurately describes the factors affecting the pressure of an ideal gas.
- Option D: Both the assertion and reason are true, not false.

Key Concepts:
- Ideal Gas Law
- Pressure
- Volume
- Temperature
- Number of Moles

Memory Aids:
- PV = nRT (remember the ideal gas law).

Quick Tips:
- Remember that the ideal gas law applies to ideal gases and provides a relationship between pressure, volume, temperature, and the number of moles.

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Application +4 | -1

1 mole of a gas having

\gamma=\frac{7}{5}

is mixed with 1 mole of a gas having

\gamma=\frac{4}{3}

. What will be the

\gamma

for the mixture?

View answer and explanation

Correct option: C

\frac{15}{11}

Option C (

\frac{15}{11}

) is correct because it correctly calculates the effective gamma for the mixture using the number of moles and the specific heat capacities.

Solution approach

Correct Answer + Core Reason: $\frac{15}{11}$ . The effective $\gamma$ for the mixture is calculated using the specific heat capacities of the individual gases.

Step-by-Step Solution:

Step 1: Define the problem. Find the $\gamma$ for the mixture of two gases with different $\gamma$ values.

Step 2: Calculate the specific heat capacities for each gas. For gas 1, $\gamma_1 = \frac{7}{5}$ . Thus, $C_{v1} = \frac{R}{{\gamma_1 - 1}} = \frac{R}{{\frac{7}{5} - 1}} = \frac{R}{{\frac{2}{5}}} = \frac{5}{2}R$ and $C_{p1} = C_{v1} + R = \frac{5}{2}R + R = \frac{7}{2}R$ .\nFor gas 2, $\gamma_2 = \frac{4}{3}$ . Thus, $C_{v2} = \frac{R}{{\gamma_2 - 1}} = \frac{R}{{\frac{4}{3} - 1}} = \frac{R}{{\frac{1}{3}}} = 3R$ and $C_{p2} = C_{v2} + R = 3R + R = 4R$ .

Step 3: Calculate the total specific heat capacities for the mixture. Since we have 1 mole of each gas, the total $C_v$ is $C_v = C_{v1} + C_{v2} = \frac{5}{2}R + 3R = \frac{5}{2}R + \frac{6}{2}R = \frac{11}{2}R$ . The total $C_p$ is $C_p = C_{p1} + C_{p2} = \frac{7}{2}R + 4R = \frac{7}{2}R + \frac{8}{2}R = \frac{15}{2}R$ .

Step 4: Calculate the $\gamma$ for the mixture. $\gamma_{mix} = \frac{{C_p}}{{C_v}} = \frac{{\frac{15}{2}R}}{{\frac{11}{2}R}} = \frac{15}{11}$ .

Why other options are wrong:
- Option A: $\frac{5}{11}$ This is incorrect because it is the inverse of the correct answer.
- Option B: $\frac{15}{13}$ This is incorrect because it doesn't correctly calculate the total specific heat capacities.
- Option D: $\frac{5}{13}$ This is incorrect because it doesn't correctly calculate the total specific heat capacities.

Key Concepts:
- Specific heat capacity
- Adiabatic index
- Mixture of gases

Memory Aids:
- $\gamma = \frac{{C_p}}{{C_v}}$
- $C_p = C_v + R$

Quick Tips:
- Remember to calculate the total specific heat capacities for the mixture before finding the effective $\gamma$ .

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Remembering +4 | -1

The ratio of two specific heats

\frac{{{C_P}}}{{{C_V}}}

of

CO

is

View answer and explanation

Correct option: B

1.4

Option B (1.4) is correct because carbon monoxide (CO) is a diatomic gas, and the ratio of specific heats for diatomic gases is approximately 1.4.

Solution approach

Correct Answer + Core Reason: 1.4. The ratio of specific heats for a diatomic gas is 1.4.

Step-by-Step Solution:

Step 1: Identify the gas. CO is a diatomic gas.

Step 2: Recall the degrees of freedom for a diatomic gas. It has 5 degrees of freedom (3 translational and 2 rotational).

Step 3: Calculate $C_V$ using the equipartition theorem: $C_V = \frac{f}{2}R = \frac{5}{2}R$ , where $f$ is the number of degrees of freedom and $R$ is the ideal gas constant.

Step 4: Calculate $C_P$ using the relation $C_P = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R$ .

Step 5: Calculate the ratio $\gamma = \frac{C_P}{C_V} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} = 1.4$ .

Why other options are wrong:
- Option A: 1.33 is approximately the ratio for a polyatomic gas.
- Option C: 1.29 is not a standard value for the ratio of specific heats.
- Option D: 1.66 is the ratio of specific heats for a monatomic gas.

Key Concepts:
- Specific Heats
- Degrees of Freedom
- Diatomic Gas

Memory Aids:
- Diatomic = 1.4.

Quick Tips:
- Remember the standard ratios for monatomic, diatomic, and polyatomic gases.

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Understanding +4 | -1

For a gas

\gamma = 7/5.

The gas may probably be

View answer and explanation

Correct option: B

Hydrogen

Option B (Hydrogen) is correct because diatomic gases like Hydrogen have a

\gamma

value of approximately 7/5 (1.4).

Solution approach

Correct Answer + Core Reason: Hydrogen. A diatomic gas has $\gamma = 7/5$ .

Step-by-Step Solution:

Step 1: Recall the definition of $\gamma$ : $\gamma = \frac{{C_p}}{{C_v}}$ , where $C_p$ is the molar specific heat at constant pressure and $C_v$ is the molar specific heat at constant volume.

Step 2: Consider the degrees of freedom for different types of gases:
- Monoatomic gas: 3 degrees of freedom (translational only), $C_v = \frac{{3}}{{2}}R$ , $C_p = \frac{{5}}{{2}}R$ , $\gamma = \frac{{5}}{{3}} \approx 1.67$
- Diatomic gas: 5 degrees of freedom (3 translational + 2 rotational), $C_v = \frac{{5}}{{2}}R$ , $C_p = \frac{{7}}{{2}}R$ , $\gamma = \frac{{7}}{{5}} = 1.4$
- Polyatomic gas: 6 degrees of freedom (3 translational + 3 rotational), $C_v = 3R$ , $C_p = 4R$ , $\gamma = \frac{{4}}{{3}} \approx 1.33$

Step 3: Analyze the given options:
- Helium: Monoatomic, $\gamma = \frac{{5}}{{3}}$
- Hydrogen: Diatomic, $\gamma = \frac{{7}}{{5}}$
- Argon: Monoatomic, $\gamma = \frac{{5}}{{3}}$
- Neon: Monoatomic, $\gamma = \frac{{5}}{{3}}$

Step 4: Conclusion: Hydrogen is the only gas with $\gamma = \frac{{7}}{{5}}$ .

Why other options are wrong:
- Option A: Helium is a monoatomic gas with $\gamma = \frac{{5}}{{3}}$ .
- Option C: Argon is a monoatomic gas with $\gamma = \frac{{5}}{{3}}$ .
- Option D: Neon is a monoatomic gas with $\gamma = \frac{{5}}{{3}}$ .

Key Concepts:
- Adiabatic index ( $\gamma$ )
- Degrees of freedom
- Molar specific heat

Memory Aids:
- Remember that diatomic gases have $\gamma = \frac{{7}}{{5}}$ .

Quick Tips:
- Know the degrees of freedom for different types of gases to determine their $\gamma$ values.

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Application +4 | -1

Average K.E. of one single molecule of monoatomic gas is

0.414 \mathrm{eV}

(of

\perp

molecule). What will be is temp

=

?

View answer and explanation

Correct option: B

3200 K

Option B (3200 K) is correct because the temperature is calculated using the average kinetic energy formula for a monoatomic gas,

KE = \frac{3}{2}kT

.

Solution approach

Correct Answer + Core Reason: 3200 K. The temperature is calculated using the average kinetic energy formula.

Step-by-Step Solution:

Step 1: Recall the formula for average kinetic energy of a monoatomic gas: $KE = \frac{3}{2}kT$

Step 2: Convert the kinetic energy from eV to Joules: $KE = 0.414 eV \times 1.6 \times 10^{-19} J/eV = 0.414 \times 1.6 \times 10^{-19} J$

Step 3: Substitute the values into the formula: $\frac{3}{2}kT = 0.414 \times 1.6 \times 10^{-19} J$

Step 4: Solve for T: $T = \frac{2 \times 0.414 \times 1.6 \times 10^{-19} J}{3 \times k} = \frac{2 \times 0.414 \times 1.6 \times 10^{-19} J}{3 \times 1.38 \times 10^{-23} J/K}$

Step 5: Calculate the temperature: $T = \frac{1.3248 \times 10^{-19}}{4.14 \times 10^{-23}} K \approx 3200 K$

Why other options are wrong:
- Option A: 3000 K: This is incorrect due to a miscalculation or rounding error.
- Option C: 4292 K: This is incorrect due to a miscalculation or incorrect substitution of values.
- Option D: 2500 K: This is incorrect due to a miscalculation or incorrect substitution of values.

Key Concepts:
- Kinetic theory of gases
- Average kinetic energy of a molecule
- Boltzmann constant

Memory Aids:
- Remember the formula: $KE = \frac{3}{2}kT$

Quick Tips:
- Always convert units to SI units before substituting into the formula.
- Be careful with the Boltzmann constant value.

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Understanding +4 | -1

Three vessels of equal capacity have gases at
the same temperature and pressure. The first
vessel contains helium (monoatomic), the
second contains fluorine (diatomic) and the
third contains sulfur hexafluoride (polyatomic).
The correct statement, among the following is:

View answer and explanation

Correct option: D

The root mean square speed of helium is
the largest

Option D (The root mean square speed of helium is the largest) is correct because RMS speed is inversely proportional to the square root of the molar mass, and helium has the smallest molar mass.

Solution approach

Correct Answer + Core Reason: The root mean square speed of helium is the largest. RMS speed is inversely proportional to the square root of molar mass.

Step-by-Step Solution:

Step 1: Write the formula for root mean square speed: $v_{rms} = \sqrt{\frac{{3RT}}{M}}$ , where $R$ is the gas constant, $T$ is the temperature, and $M$ is the molar mass.

Step 2: Analyze the relationship between $v_{rms}$ and $M$ : Since $R$ and $T$ are the same for all vessels, $v_{rms} \propto \frac{1}{{\sqrt{M}}}$ .

Step 3: Compare the molar masses of the gases: Helium (He) has the smallest molar mass among the given gases (Helium, Fluorine, and Sulfur Hexafluoride).

Step 4: Conclude that helium has the largest $v_{rms}$ : Because helium has the smallest molar mass, it will have the largest root mean square speed.

Why other options are wrong:
- Option A: Sulfur hexafluoride has the largest molar mass, so it has the smallest RMS speed.
- Option B: Since the temperature and pressure are the same, and the volumes are equal, the number of molecules is the same according to Avogadro's law.
- Option C: The RMS speed depends on the molar mass, so it is not the same for all three gases.

Key Concepts:
- Root Mean Square Speed
- Kinetic Theory of Gases
- Molar Mass
- Avogadro's Law

Memory Aids:
- RMS speed is inversely proportional to the square root of Molar mass.

Quick Tips:
- Remember the formula for RMS speed and its dependence on molar mass.

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Understanding +4 | -1

In a dilute gas at pressure

p

and temperature

T,

the mean time between successive collisions of a molecule varies with

T

as

View answer and explanation

Correct option: B

\frac{1}{{\sqrt T }}

Option B is correct because the mean time between collisions is inversely proportional to the square root of the temperature.

Solution approach

Correct Answer + Core Reason: $\frac{{1}}{{\sqrt T }}$ . The mean time between collisions is inversely proportional to the square root of temperature.

Step-by-Step Solution:

Step 1: Recall the formula for the mean time between successive collisions: $\tau = \frac{{\lambda}}{{\upsilon_{avg}}}$

Step 2: Recall the dependence of mean free path on temperature: $\lambda \propto T^0$ (approximately constant)

Step 3: Recall the dependence of average speed on temperature: $\upsilon_{avg} \propto \sqrt{T}$

Step 4: Combine the relationships: $\tau = \frac{{\lambda}}{{\upsilon_{avg}}} \propto \frac{{1}}{{\sqrt{T}}}$

Why other options are wrong:
- Option A: This is incorrect because the mean time is inversely proportional to the square root of temperature.
- Option C: This is incorrect because the mean time is inversely proportional to the square root of temperature.
- Option D: This is incorrect because the mean time is inversely proportional to the square root of temperature.

Key Concepts:
- Mean free path
- Average speed
- Kinetic theory of gases

Memory Aids:
- $\upsilon_{avg} \propto \sqrt{T}$

Quick Tips:
- Remember the relationships between temperature, average speed, and mean free path.

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Understanding +4 | -1

Assertion: In case of collision of gas molecules in a given amount of gas, total kinetic energy is conserved.
Reason: All collisions of the gas molecules in a given amount of gas are elastic.

View answer and explanation

Correct option: A

If both assertion and reason are true and reason is the correct explanation of assertion

Option A is correct because in an elastic collision, both momentum and kinetic energy are conserved, and the conservation of kinetic energy directly implies that the total kinetic energy remains constant.

Solution approach

Correct Answer + Core Reason: If both assertion and reason are true and reason is the correct explanation of assertion. The assertion is true because collisions are elastic, and the reason correctly explains why the assertion is true.

Step-by-Step Solution:

Step 1: Define elastic collision: An elastic collision is one in which both momentum and kinetic energy are conserved.

Step 2: Explain why total kinetic energy is conserved: If all collisions are elastic, then by definition, the total kinetic energy of the system remains constant before and after the collisions.

Step 3: Relate the reason to the assertion: The reason (all collisions are elastic) directly explains why the assertion (total kinetic energy is conserved) is true.

Why other options are wrong:
- Option B: The reason is the correct explanation of the assertion.
- Option C: The assertion is true, and the reason is also true.
- Option D: Both the assertion and reason are true.

Key Concepts:
- Elastic collision
- Conservation of kinetic energy
- Conservation of momentum

Memory Aids:
- Elastic collision = Kinetic energy conserved

Quick Tips:
- Remember that in elastic collisions, both momentum and kinetic energy are conserved.

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Remembering +4 | -1

The number of translatioaonal degrees of freedom of a diatomic gas is

View answer and explanation

Correct option: A

3

Option A (3) is correct because a diatomic gas can move along the x, y, and z axes, giving it 3 translational degrees of freedom.

Solution approach

Correct Answer + Core Reason: 3. A diatomic gas has 3 translational degrees of freedom.

Step-by-Step Solution:

Step 1: The question asks for the number of translational degrees of freedom of a diatomic gas.

Step 2: Recall the definition of degrees of freedom and translational degrees of freedom.

Step 3: Translational degrees of freedom refer to the number of independent directions in which a molecule can move freely in space.

Step 4: A diatomic gas molecule can move along the x, y, and z axes. Therefore, it has 3 translational degrees of freedom.

Why other options are wrong:
- Option B: 5 is incorrect. It includes rotational degrees of freedom as well.
- Option C: 7 is incorrect.
- Option D: 6 is incorrect.

Key Concepts:
- Degrees of freedom
- Translational degrees of freedom
- Diatomic gas

Memory Aids:
- Remember that a diatomic gas has 3 translational degrees of freedom.

Quick Tips:
- Know the degrees of freedom for different types of molecules.

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Application +4 | -1

A vessel has

6 \mathrm{~g}

of hydrogen at pressure

P

and temperature

500 \mathrm{~K}

. A small hole is made in it so that hydrogen leaks out. How much hydrogen leaks out if the final pressure is

\frac{P}{3}

and temperature falls to 300 K?

View answer and explanation

Correct option: C

2.66 g

Option C (2.66 g) is correct because the mass of gas remaining is calculated using the ideal gas law, and the difference between the initial and final mass gives the mass leaked out.

Solution approach

Correct Answer + Core Reason: 2.66 g. The mass leaked out is the difference between the initial and final mass, calculated using the ideal gas law.

Step-by-Step Solution:

Step 1: Write the ideal gas equation: $PV = \frac{m}{M}RT$ , where $P$ is pressure, $V$ is volume, $m$ is mass, $M$ is molar mass, $R$ is the gas constant, and $T$ is temperature.

Step 2: Since $V$ , $M$ , and $R$ are constant, we have $m \propto \frac{P}{T}$ .

Step 3: Let $m_1$ be the initial mass and $m_2$ be the final mass. Then $\frac{m_1}{m_2} = \frac{P_1/T_1}{P_2/T_2} = \frac{P_1}{T_1} \cdot \frac{T_2}{P_2}$ .

Step 4: Given $m_1 = 6\,g$ , $P_1 = P$ , $T_1 = 500\,K$ , $P_2 = \frac{P}{3}$ , and $T_2 = 300\,K$ .

Step 5: Substitute the values: $\frac{6}{m_2} = \frac{P}{500} \cdot \frac{300}{P/3} = \frac{P}{500} \cdot \frac{900}{P} = \frac{9}{5}$ .

Step 6: Solve for $m_2$ : $m_2 = \frac{6 \times 5}{9} = \frac{30}{9} = 3.33\,g \approx 3.34\,g$ .

Step 7: Calculate the mass leaked out: $\Delta m = m_1 - m_2 = 6 - 3.34 = 2.66\,g$ .

Why other options are wrong:
- Option A (3.34 g): This is the mass of hydrogen remaining, not the mass leaked out.
- Option B (3.45 g): This is incorrect due to a calculation error.
- Option D (6 g): This is the initial mass, not the mass leaked out.

Key Concepts:
- Ideal Gas Law

Memory Aids:
- Remember the ideal gas equation: $PV = nRT$ or $PV = \frac{m}{M}RT$ .

Quick Tips:
- Identify the constant variables and simplify the equation accordingly.

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Analyzing +4 | -1

Read the given statements and decide which is/are correct on the basis of kinetic theory of gases
(I) Energy of one molecule at absolute temperature is zero
(II)

r.m.s.

speeds of different gases are same at same temperature
(III) For one gram of all ideal gas kinetic energy is same at same temperature
(IV) For one mole of all ideal gases mean kinetic energy is same at same temperature

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because all the given statements are incorrect based on the kinetic theory of gases.

Solution approach

Correct Answer + Core Reason: None of these. All the given statements are incorrect according to the kinetic theory of gases.

Step-by-Step Solution:

Step 1: Analyze statement (I): Energy of one molecule at absolute temperature is zero.

Step 2: Key Concept: According to the kinetic theory, even at absolute zero, molecules possess zero-point energy, so their energy is not zero.

Step 3: Analyze statement (II): r.m.s. speeds of different gases are the same at the same temperature.

Step 4: Key Concept: The r.m.s. speed is given by $v_{rms} = \sqrt{\frac{{3RT}}{M}}$ , where M is the molar mass. Different gases have different molar masses, so their r.m.s. speeds will be different at the same temperature.

Step 5: Analyze statement (III): For one gram of all ideal gas kinetic energy is the same at the same temperature.

Step 6: Key Concept: Kinetic energy is given by $KE = \frac{{f}}{2}nRT$ , where n is the number of moles. For one gram, the number of moles will be different for different gases, so their kinetic energies will be different.

Step 7: Analyze statement (IV): For one mole of all ideal gases mean kinetic energy is the same at the same temperature.

Step 8: Key Concept: The mean kinetic energy per mole is given by $E = \frac{{f}}{2}RT$ . The degree of freedom (f) is not the same for all gases, so the kinetic energy is not the same.

Step 9: Conclusion: All statements are incorrect.

Why other options are wrong:
- Option A: All are correct: This is wrong because all statements are incorrect.
- Option B: I and IV are correct: This is wrong because both statements I and IV are incorrect.
- Option C: IV is correct: This is wrong because statement IV is incorrect.

Key Concepts:
- Kinetic theory of gases
- Root mean square speed
- Kinetic energy of gases
- Degrees of freedom

Memory Aids:
- $v_{rms} = \sqrt{\frac{{3RT}}{M}}$
- $E = \frac{{f}}{2}RT$

Quick Tips:
- Remember the formula for r.m.s speed and kinetic energy to quickly analyze such statements.
- Consider the dependence on molar mass and degrees of freedom.

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Remembering +4 | -1

At a given temperature the

r.m.s.

velocity of molecules of the gas is

View answer and explanation

Correct option: D

Inversely proportional to square root of molecular weight

Option D is correct because the root-mean-square velocity of gas molecules is inversely proportional to the square root of the molecular weight.

Solution approach

Correct Answer + Core Reason: Inversely proportional to square root of molecular weight. The RMS velocity is inversely proportional to the square root of the molar mass.

Step-by-Step Solution:

Step 1: Recall the formula for root-mean-square velocity:
$v_{rms} = \sqrt{\frac{{3RT}}{M}}$ \nwhere $v_{rms}$ is the root-mean-square velocity, $R$ is the ideal gas constant, $T$ is the temperature, and $M$ is the molar mass.

Step 2: Analyze the relationship between $v_{rms}$ and $M$ at a given temperature:\nSince $R$ and $T$ are constant, $v_{rms} \propto \frac{1}{\sqrt{M}}$

Step 3: Conclusion: The root-mean-square velocity is inversely proportional to the square root of the molecular weight.

Why other options are wrong:
- Option A: Same - This is incorrect because the RMS velocity depends on the molecular weight.
- Option B: Proportional to molecular weight - This is incorrect because the RMS velocity is inversely proportional to the square root of the molecular weight.
- Option C: Inversely proportional to molecular weight - This is incorrect because the RMS velocity is inversely proportional to the *square root* of the molecular weight, not the molecular weight itself.

Key Concepts:
- Root-mean-square velocity
- Kinetic theory of gases

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

Quick Tips:
- Always remember the correct formula for RMS velocity and the relationship between RMS velocity and molecular weight.

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Analyzing +4 | -1

A certain ideal gas undergoes a polytropic process

p V^n=

constant such that the molar specific heat during the process is negative. If the ratio of the specific heat of the gas be

\gamma

, then the range of values of

n

will be

View answer and explanation

Correct option: B

1 < n < \gamma

Option B is correct because the molar specific heat is negative when 1 < n < gamma, as derived from the polytropic process equation.

Solution approach

Correct Answer + Core Reason: $1 < n < \gamma$ . The range of n is derived from the condition that the molar specific heat is negative in a polytropic process.

Step-by-Step Solution:

Step 1: Given the polytropic process $pV^n = \text{{constant}}$ and the ideal gas law $pV = RT$ (for 1 mole).

Step 2: The molar specific heat in a polytropic process is given by: $C = C_V + \frac{{R}}{{1 - n}} = \frac{{R}}{{\gamma - 1}} + \frac{{R}}{{1 - n}}$ .

Step 3: Simplify the expression: $C = R \left( {\frac{{1}}{{\gamma - 1}} + \frac{{1}}{{1 - n}}} \right) = R \left( {\frac{{1 - n + \gamma - 1}}{{(\gamma - 1)(1 - n)}}} \right) = R \frac{{\gamma - n}}{{(\gamma - 1)(1 - n)}}$ .

Step 4: For $C$ to be negative, we need $\frac{{\gamma - n}}{{(1 - n)}} < 0$ since $R$ and $(\gamma - 1)$ are positive.

Step 5: This inequality holds when $1 < n < \gamma$ .

Why other options are wrong:
- Option A: $0 < n < \gamma$ : This is incorrect because it doesn't satisfy the condition for negative molar specific heat.
- Option C: $n = \gamma$ : This would make the molar specific heat zero, not negative.
- Option D: $n > \gamma$ : This would result in a positive molar specific heat.

Key Concepts:
- Polytropic process
- Molar specific heat
- Ideal gas law

Memory Aids:
- Remember the formula for molar specific heat in a polytropic process.

Quick Tips:
- Pay attention to the sign conventions when dealing with inequalities.

210 Medium

210 of 257

Analyzing +4 | -1

The specific heats,

{C_p}\mathrm{ \,\,and }\,\,{C_V}

of a gas of diatomic molecules, A are given (in units of

\mathrm{J mo}{\mathrm{l}^{ - 1}}{K^{ - 1}}

) by 29 and 22, respectively. Another gas of diatomic molecules B, has the corresponding values 30 and 21. If they are treated as ideal gases, then

View answer and explanation

Correct option: A

A has a vibrational mode but B has none

Option A (A has a vibrational mode but B has none) is correct because the calculated degrees of freedom for gas A is approximately 7, indicating a vibrational mode, while for gas B it is 5, indicating no vibrational mode.

Solution approach

Correct Answer + Core Reason: A has a vibrational mode but B has none. The degrees of freedom indicate the presence or absence of vibrational modes.

Step-by-Step Solution:

Step 1: For a diatomic gas, the relationship between specific heats is given by $\gamma = \frac{{C_p}}{{C_V}} = 1 + \frac{{2}}{{f}}$ , where $f$ is the number of degrees of freedom.

Step 2: For gas A, $C_p = 29$ and $C_V = 22$ . Therefore, $\gamma_A = \frac{{29}}{{22}} \approx 1.318$ .

Step 3: Using the formula, $1.318 = 1 + \frac{{2}}{{f_A}}$ , so $\frac{{2}}{{f_A}} = 0.318$ , and $f_A = \frac{{2}}{{0.318}} \approx 6.29 \approx 7$ .

Step 4: For gas B, $C_p = 30$ and $C_V = 21$ . Therefore, $\gamma_B = \frac{{30}}{{21}} \approx 1.429$ .

Step 5: Using the formula, $1.429 = 1 + \frac{{2}}{{f_B}}$ , so $\frac{{2}}{{f_B}} = 0.429$ , and $f_B = \frac{{2}}{{0.429}} \approx 4.66 \approx 5$ .

Step 6: A diatomic molecule has 5 degrees of freedom (3 translational and 2 rotational) without vibrational modes. Gas A has approximately 7 degrees of freedom, indicating a vibrational mode, while gas B has approximately 5 degrees of freedom, indicating no vibrational mode.

Why other options are wrong:
- Option B: Both A and B have a vibrational mode each is incorrect because gas B has 5 degrees of freedom, indicating no vibrational mode.
- Option C: A has one vibrational mode and B has two is incorrect because gas B has 5 degrees of freedom, indicating no vibrational mode.
- Option D: A is right but B has a vibrational mode is incorrect because gas B has 5 degrees of freedom, indicating no vibrational mode.

Key Concepts:
- Specific heat
- Degrees of freedom
- Diatomic molecules

Memory Aids:
- Remember that a diatomic molecule has 5 degrees of freedom (3 translational and 2 rotational) without vibrational modes.

Quick Tips:
- Calculate the degrees of freedom to determine the presence of vibrational modes.

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