JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 8)
A vessel contains $$14 \mathrm{~g}$$ of nitrogen gas at a temperature of $$27^{\circ} \mathrm{C}$$. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be :
Take $$\mathrm{R}=8.32 \mathrm{~J} \mathrm{~mol}^{-1} \,\mathrm{k}^{-1}$$.
2229 J
5616 J
9360 J
13,104 J
Explanation
n = 0.5
T = 300
For vrms to be doubled T' = 4 $$\times$$ 300 = 1200
$$\Rightarrow$$ Heat transferred
$$ = (0.5)\left( {{5 \over 2}} \right)(8.32)(900)$$
$$ = 9360$$ J
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