Question

A vessel contains \(14 g\) of nitrogen gas at a temperature of \(27°C\). The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be: 

Take \(R=8.32 J-mol^{-1}K^{-1}\)

• A. 2229 J
• B. 5616 J
• C. 9360 J
• D. 13,104 J

Answer 1

The Correct Option is C

To solve this problem, we need to calculate the amount of heat required to double the root mean square (r.m.s) speed of nitrogen gas molecules in the vessel. Let's go through the solution step-by-step.

Step 1: Understand the relationship between r.m.s speed and temperature

The r.m.s speed (v_{\text{rms}}) of gas molecules is given by the formula:

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

where R is the universal gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.

Step 2: Calculate the temperature to double the r.m.s speed

If we want to double the r.m.s speed, let's denote the initial r.m.s speed as v_1 and the final r.m.s speed as v_2 = 2v_1. Using the r.m.s speed formula:

v_2 = \sqrt{\frac{3RT_2}{M}} = 2 \times v_1 = 2 \times \sqrt{\frac{3RT_1}{M}}

Squaring both sides, we get:

\frac{3RT_2}{M} = 4 \times \frac{3RT_1}{M}

This implies:

T_2 = 4 \times T_1

Initially, T_1 = 27^\circ C = 300 K. Therefore, T_2 = 4 \times 300 = 1200 K.

Step 3: Calculate the amount of heat required

The change in internal energy of the gas is given by the formula:

\Delta U = nC_v\Delta T

Here, C_v for nitrogen, a diatomic gas, is \frac{5}{2}R. Therefore, the change in internal energy is:

\Delta U = n \times \frac{5}{2}R \times (T_2 - T_1)

The molar mass of nitrogen, N_2, is 28 \, g\,mol^{-1}. Thus, the number of moles n is:

n = \frac{14}{28} = 0.5 \, \text{mol}

Substituting values:

\Delta U = 0.5 \times \frac{5}{2} \times 8.32 \times (1200 - 300)

\Delta U = 2.5 \times 8.32 \times 900

\Delta U = 5 \times 748.8 \, J = 9360 \, J

Thus, the correct answer is 9360 J.