Question

An ideal gas in a closed container is heated so that the final rms speed of the gas particles increases by $2$ times the initial rms speed. If the initial gas temperature is $27^{\circ} C$, then the final temperature of the ideal gas is :

• A. $1200^{\circ} C$
• B. $927^{\circ} C$
• C. $827^{\circ} C$
• D. $1473^{\circ} C$

Answer 1

Answer: (B)

As, rms velocity of ideal gas particles is given,
$v_{ rms }=\sqrt{\frac{3 k T}{m}}$
$v_{ rms } \propto \sqrt{T}$
where, temperature $T$ is measured in kelvin scale. So, the ratio
$\frac{v_{1 rms }^{2}}{v_{2 rms }^{2}}=\frac{T_{1}}{T_{2}}$
As given, $v_{2 rms }=2 v_{1 rms }$ and $T_{1}=27^{\circ} C =300 \,K$
$\Rightarrow \, \frac{v_{1 rms }^{2}}{4 v_{1 rms }^{2}}=\frac{300}{x}$
$\Rightarrow \, x=300 \times 4=1200\, K$
Hence, $1200 K =(1200-273){ }^{\circ} C =927^{\circ} C$