Kinetic Theory of Gases Previous Year Questions — JEE Main Physics
26 past-year Kinetic Theory of Gases questions from JEE Main (Physics) — page 4.
Q50 (2024)
Two vessels A and B are of the same size and are at same temperature. A contains $1 \mathrm{~g}$ of hydrogen and B contains $1 \mathrm{~g}$ of oxygen. $P_A$ and $P_B$ are the pressures of the gases in A and B respectively, then $\frac{P_A}{P_B}$ is:
- 16
- 8
- 4
- 32
### Related Formula
From the Ideal Gas Law equation:
$$P V = n R T \implies P = \frac{n R T}{V}$$
### Core Logic
Given that both vessels possess the same volume ($V_A = V_B$) and identical temperature states ($T_A = T_B$), the ratio of pressure reduces directly to:
$$\frac{P_A}{P_B} = \frac{n_A}{n_B}$$
where $n_A$ and $n_B$ are the number of moles of Hydrogen and Oxygen respectively.
### Step 1: Calculate the Number of Moles
For Hydrogen ($H_2$, molar mass = $2 \mathrm{~g/mol}$):
$$n_A = \frac{1}{2}$$
For Oxygen ($O_2$, molar mass = $32 \mathrm{~g/mol}$):
$$n_B = \frac{1}{32}$$
### Step 2: Find the Pressure Ratio
Substituting mole counts into the direct ratio:
$$\frac{P_A}{P_B} = \frac{1/2}{1/32} = \frac{32}{2} = 16$$
Therefore, the pressure ratio $\frac{P_A}{P_B}$ is $16$.
### Pattern Recognition
For gas mixtures or vessel comparisons under constant volume and temperature, pressure matches the molar abundance directly ($P \propto n$). Remember that standard elementary gases ($H_2, O_2, N_2$) exist as diatomic configurations when defining molar values.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q46 (2024)
If three moles of monoatomic gas $\left(\gamma = \frac{5}{3}\right)$ is mixed with two moles of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$, the value of adiabatic exponent $\gamma$ for the mixture is:
- $1.75$
- $1.40$
- $1.52$
- $1.35$
### Related Formula
$$f_{\text{mixture}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$$
$$\gamma_{\text{mixture}} = 1 + \frac{2}{f_{\text{mixture}}}$$
### Core Logic
For a monoatomic gas, degrees of freedom $f_1 = 3$. Moles $n_1 = 3$.
For a diatomic gas, degrees of freedom $f_2 = 5$. Moles $n_2 = 2$.
We can compute the equivalent degrees of freedom for the mixture using a weighted average.
### Step 1: Calculate Equivalent Degrees of Freedom
$$f_{\text{mixture}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$$
$$f_{\text{mixture}} = \frac{3(3) + 2(5)}{3 + 2} = \frac{9 + 10}{5} = \frac{19}{5}$$
### Step 2: Calculate Adiabatic Exponent
$$\gamma_{\text{mixture}} = 1 + \frac{2}{f_{\text{mixture}}}$$
$$\gamma_{\text{mixture}} = 1 + \frac{2}{\frac{19}{5}} = 1 + \frac{10}{19} = \frac{29}{19} \approx 1.52$$
### Pattern Recognition
Alternatively, you can compute $C_v$ and $C_p$ for the mixture: $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$, and $\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}}$. Both methods yield identical results rapidly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Class 11 Physics: Thermodynamics
Q49 (2024)
At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at $47^{\circ}\mathrm{C}$?
- $80 \mathrm{~K}$
- $-73 \mathrm{~K}$
- $4 \mathrm{~K}$
- $20 \mathrm{~K}$
### Related Formula
$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$
### Core Logic
For the RMS velocities to be equal, the ratio of temperature to molar mass ($T/M$) must be identical for both gases.
### Step 1: Set Up Equivalency
$$\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$$
$$\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$$
### Step 2: Substitute Values
$T_{O_2} = 47^\circ\mathrm{C} = 47 + 273 = 320 \mathrm{~K}$
$M_{H_2} = 2 \mathrm{~g/mol}$
$M_{O_2} = 32 \mathrm{~g/mol}$
$$\frac{T_{H_2}}{2} = \frac{320}{32}$$
$$T_{H_2} = 2 \times 10 = 20 \mathrm{~K}$$
### Pattern Recognition
$v_{\text{rms}}$ scales strictly as $\sqrt{T/M}$. Remember to always convert Celsius to Kelvin before substituting.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q38 (2024)
A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is
- $29 \text{ RT}$
- $20 \text{ RT}$
- $27 \text{ RT}$
- $21 \text{ RT}$
### Related Formula
$U = n C_v T$
where $C_v = \frac{f}{2} R$ and $f$ is the degree of freedom.
### Core Logic
Argon (Ar) is monatomic $\implies f_1 = 3 \implies C_{v_1} = \frac{3R}{2}$.
Oxygen ($O_2$) is diatomic $\implies f_2 = 5$ (neglecting vibrational modes) $\implies C_{v_2} = \frac{5R}{2}$.
Total internal energy $U = U_1 + U_2 = n_1 C_{v_1} T + n_2 C_{v_2} T$.
### Step 1: Compute Total Energy
$$U = 8 \times \left(\frac{3R}{2}\right) T + 6 \times \left(\frac{5R}{2}\right) T$$
$$U = 4(3RT) + 3(5RT)$$
$U = 12RT + 15RT$
$U = 27 RT$
### Pattern Recognition
Internal energy is strictly additive. Immediately map Monatomic $\to 3/2$ and Diatomic $\to 5/2$. Plug linearly: $8(1.5) + 6(2.5) = 12 + 15 = 27$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Class 11 Physics: Thermodynamics
Q31 (2024)
The parameter that remains the same for molecules of all gases at a given temperature is :
- $\text{kinetic energy}$
- $\text{momentum}$
- $\text{mass}$
- $\text{speed}$
### Related Formula
$$\text{KE} = \frac{f}{2}kT$$
### Core Logic
The average translational kinetic energy of any gas molecule depends only on the absolute temperature of the gas and is independent of the nature or mass of the gas.
For 1 mole of any ideal gas, the average translational kinetic energy is $\frac{3}{2}RT$. Therefore, at a given temperature, the kinetic energy parameter is uniform across all ideal gases.
### Pattern Recognition
Temperature is directly proportional to average translational kinetic energy. If T is constant, KE is constant for all gases regardless of mass.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory
