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 $\mathrm{R}=8.32 \mathrm{~J} \mathrm{~mol}^{-1} \,\mathrm{k}^{-1}$.
Solution
<p>n = 0.5</p>
<p>T = 300</p>
<p>For v<sub>rms</sub> to be doubled T' = 4 $\times$ 300 = 1200</p>
<p>$\Rightarrow$ Heat transferred</p>
<p>$= (0.5)\left( {{5 \over 2}} \right)(8.32)(900)$</p>
<p>$= 9360$ J</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
This question is part of PrepWiser's free JEE Main question bank. 271 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.