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

Answer: (C)

9360 J

Answer 2

$v_{rms} ∝ \sqrt {T}v_{rms} ∝ \sqrt{300}K$, $v_{rms_f} = 2v_{rms_i}$

$T_f = 1200K$, $T_i = 300K$,

$n= \frac{14}{28}$ $= \frac{1}{2}$

$Q= nC_v∆T$ $=\frac{1}{2} × \frac{5R}{2 }× 900$

$Q = 9360 J$