A vessel contains 14 ~g of nitrogen gas at a temperature of 27^\circ 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 .

Heat and Thermodynamics PYQ — JEE Main 2022 · 28 Jul · Shift 2 · Q47 with Solution

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Correct answer: C

Given data
- Mass of nitrogen gas: $m = 14\,\text{g}$
- Molar mass of $N_2$ : $M = 28\,\text{g mol}^{-1}$
- Initial temperature: $T_1 = 27^\circ\text{C} = 300\,\text{K}$
- Gas constant: $R = 8.32\,\text{J mol}^{-1}\text{K}^{-1}$
Number of moles $n = \frac{m}{M} = \frac{14}{28} = 0.5\,\text{mol}$
Relation between r.m.s. speed and temperature For an ideal gas, $v_{\text{rms}} \propto \sqrt{T}$ If the r.m.s. speed is doubled, $\frac{v_2}{v_1} = 2 = \sqrt{\frac{T_2}{T_1}}$ Squaring both sides, $\frac{T_2}{T_1} = 4$ So, $T_2 = 4T_1 = 4 \times 300 = 1200\,\text{K}$
Temperature rise $\Delta T = T_2 - T_1 = 1200 - 300 = 900\,\text{K}$
Heat supplied Since the gas is in a vessel (rigid container), the process is at constant volume. Hence, $Q = nC_V\Delta T$

For diatomic gas ( $N_2$ ), neglecting vibration, $C_V = \frac{5R}{2}$

Therefore, $Q = n\left(\frac{5R}{2}\right)\Delta T$
Substitute values $Q = 0.5 \times \frac{5}{2} \times 8.32 \times 900$

First, $\frac{5}{2} \times 8.32 = 20.8$

Then, $Q = 0.5 \times 20.8 \times 900$ $Q = 10.4 \times 900$ $Q = 9360\,\text{J}$
Option check
- A: $2229\,\text{J}$
- B: $5616\,\text{J}$
- C: $9360\,\text{J}$ ✅
- D: $13104\,\text{J}$

Therefore, the correct answer is Option C.