Kinetic Theory of Gases Previous Year Questions — JEE Main Physics
26 past-year Kinetic Theory of Gases questions from JEE Main (Physics) — page 3.
Q1297 (2025)
A container of fixed volume contains a gas at $27^{\circ}\mathrm{C}$ . To double the pressure of the gas, the temperature of gas should be raised to _________ $^{\circ}\mathrm{C}$
### Related Formula
$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$
### Core Logic
Initial temperature $T_1 = 27 + 273 = 300\text{ K}$. Since volume is kept fixed :
$$\frac{P}{300} = \frac{2P}{T_2} \implies T_2 = 600\text{ K}$$
Converting back to Celsius :
$$T_2 = 600 - 273 = 327^{\circ}\mathrm{C}$$
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q36 (2024)
Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is:
- $\frac{9}{4} R$
- $\frac{7}{4} R$
- $\frac{3}{2} R$
- $\frac{5}{2} R$
### Related Formula
Molar specific heat at constant volume for a gas mixture:
$$C_{V\text{, mix}} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$$
For a monoatomic gas:
$$C_{V1} = \frac{3}{2}R$$
For a diatomic gas:
$$C_{V2} = \frac{5}{2}R$$
### Core Logic
Given values:
$n_1 = 2$ (monoatomic), $n_2 = 6$ (diatomic).
Substitute these inputs directly into the mixture equation:
$$C_{V\text{, mix}} = \frac{2 \times \left(\frac{3}{2}R\right) + 6 \times \left(\frac{5}{2}R\right)}{2 + 6}$$
### Step 1: Simplify Expression
$$C_{V\text{, mix}} = \frac{3R + 15R}{8} = \frac{18R}{8} = \frac{9}{4}R$$
### Pattern Recognition
Weighted average rule based on internal degrees of freedom: total internal energy changes scale additively with mole numbers.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Class 11 Physics: Thermodynamics
Q35 (2024)
The temperature of a gas having $2.0 \times 10^{25}$ molecules per cubic meter at $1.38\text{ atm}$ (Given, $k = 1.38 \times 10^{-23}\text{ J K}^{-1}$) is:
- $500\text{ K}$
- $200\text{ K}$
- $100\text{ K}$
- $300\text{ K}$
### Related Formula
The state equation of an ideal gas in terms of the number of molecules $N$ and Boltzmann constant $k$ is:
$PV = NkT$
Rearranging to express pressure in terms of number density $n = N/V$:
$P = n k T$
### Core Logic
Given parameters:
* Number density, $n = \frac{N}{V} = 2.0 \times 10^{25}\text{ molecules/m}^3$
* Pressure, $P = 1.38\text{ atm} = 1.38 \times 1.01 \times 10^5\text{ N/m}^2$
* Boltzmann constant, $k = 1.38 \times 10^{-23}\text{ J K}^{-1}$
### Step 1: Solve for Temperature
Rearranging $P = n k T$ for temperature $T$:
$$T = \frac{P}{nk}$$
Substitute the values:
$$T = \frac{1.38 \times 1.01 \times 10^5}{(2.0 \times 10^{25}) \times (1.38 \times 10^{-23})}$$
Notice that the term $1.38$ cancels out from numerator and denominator:
$$T = \frac{1.01 \times 10^5}{2.0 \times 10^2}$$
$$T = \frac{1.01 \times 10^3}{2.0} \approx \frac{1010}{2} \approx 500\text{ K}$$
### Pattern Recognition
The numerical values are designed to cancel out smoothly. Spotting the $1.38$ cancelation instantly saves valuable calculation time.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q48 (2024)
$N$ moles of a polyatomic gas ($f = 6$) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of $N$ is:
- $6$
- $3$
- $4$
- $2$
### Related Formula
The equivalent degrees of freedom $f_{\text{eq}}$ for a mixture of gases is:
$$f_{\text{eq}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$$
where:
* $n_1, n_2$ are the number of moles of each gas.
* $f_1, f_2$ are the respective degrees of freedom of each gas.
### Core Logic
For the given gases:
1. Polyatomic gas:
* Moles, $n_1 = N$
* Degrees of freedom, $f_1 = 6$
2. Monoatomic gas:
* Moles, $n_2 = 2$
* Degrees of freedom, $f_2 = 3$
We want the mixture to behave as a diatomic gas. For a diatomic gas:
* Equivalent degrees of freedom, $f_{\text{eq}} = 5$
### Step 1: Solve for N
Substitute the values into the degrees of freedom mixture formula:
$$5 = \frac{(N)(6) + (2)(3)}{N + 2}$$
$$5(N + 2) = 6N + 6$$
$$5N + 10 = 6N + 6$$
$$10 - 6 = 6N - 5N \implies N = 4$$
### Pattern Recognition
Diatomic equivalent degree of freedom is 5. Since the monoatomic degrees of freedom (3) and polyatomic degrees of freedom (6) bracket 5, you can use the weighted ratio method to find the molar proportions directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q50 (2024)
The average kinetic energy of a monatomic molecule is $0.414\text{ eV}$ at temperature:
- $3000\text{ K}$
- $3200\text{ K}$
- $1600\text{ K}$
- $1500\text{ K}$
### Related Formula
$$K_{\text{avg}} = \frac{3}{2} k_B T$$
### Core Logic
Given energy is in electron-volts ($1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$), we isolate $T$:
$$T = \frac{2 K_{\text{avg}}}{3 k_B}$$
Substitute constants ($k_B = 1.38 \times 10^{-23}\text{ J/K}$):
### Step 1: Compute value
$$T = \frac{2 \times 0.414 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}}$$
$$T = \frac{1.3248 \times 10^{-19}}{4.14 \times 10^{-23}} = 0.32 \times 10^4 = 3200\text{ K}$$
### Pattern Recognition
Converting eV energy properties straight to structural SI standard Joules reveals highly cleanly simplified scalar components when paired with Boltzmann values.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
