Question

A mixture of 4 gm helium and 28 gm of nitrogen is enclosed in a vessel of constant volume at 300K. The quantity of heat absorbed by the mixture to increase root mean velocity of its molecules by 50% is (R is universal gas constant)

• A. 1500 R
• B. 450 R

Answer 1

Answer: (A)

1500 R

Answer 2

1. Temperature Change: The root mean square (RMS) velocity of gas molecules is directly proportional to the square root of the absolute temperature ($v_{rms} \propto \sqrt{T}$). If the RMS velocity increases by 50%, the new velocity is $1.5$ times the original velocity. Therefore, the new temperature $T_2$ is related to the initial temperature $T_1$ by $T_2 = (1.5)^2 T_1 = 2.25 T_1$. Given $T_1 = 300$ K, the final temperature is $T_2 = 2.25 \times 300 \text{ K} = 675 \text{ K}$. The temperature change is $\Delta T = T_2 - T_1 = 675 \text{ K} - 300 \text{ K} = 375 \text{ K}$.

2. Number of Moles:

   • For Helium (He, Molar mass = 4 g/mol): $n_{He} = \frac{4 \text{ gm}}{4 \text{ g/mol}} = 1$ mol.
   • For Nitrogen ($N_2$, Molar mass = 28 g/mol): $n_{N_2} = \frac{28 \text{ gm}}{28 \text{ g/mol}} = 1$ mol.
   • The total number of moles in the mixture is $n_{total} = n_{He} + n_{N_2} = 1 + 1 = 2$ mol.

3. Molar Heat Capacity of the Mixture ($C_{v, mix}$):

   • Helium is a monatomic gas, so its molar heat capacity at constant volume is $C_{v, He} = \frac{3}{2}R$.
   • Nitrogen is a diatomic gas, so its molar heat capacity at constant volume is $C_{v, N_2} = \frac{5}{2}R$.
   • The molar heat capacity of the mixture at constant volume is calculated as: $C_{v, mix} = \frac{n_{He}C_{v, He} + n_{N_2}C_{v, N_2}}{n_{total}} = \frac{(1 \text{ mol}) \times (\frac{3}{2}R) + (1 \text{ mol}) \times (\frac{5}{2}R)}{2 \text{ mol}} = \frac{\frac{3}{2}R + \frac{5}{2}R}{2} = \frac{4R}{2} = 2R$.

4. Heat Absorbed ($Q$): Since the vessel has a constant volume, the heat absorbed is given by $Q = n_{total} C_{v, mix} \Delta T$. $Q = (2 \text{ mol}) \times (2R) \times (375 \text{ K}) = 4R \times 375 = 1500R$.