Question

The temperature of a gas is raised from ${27}^{\circ }\text{C}$ to ${927}^{\circ }\text{C}$. The root mean square speed

• A. becomes $\sqrt{\frac{927}{27}}$ times the earlier value.
• B. remains the same.
• C. gets halved.
• D. gets doubled.

Answer 1

The correct option is D gets doubled.

Given,
Initial temperature, $T_1={27}^{\circ }\text{C}=300\text{K}$
Final temperature, $T_2={927}^{\circ }\text{C}=1200\text{K}$
We know that,
$v_{rms}=\sqrt{\frac{3RT}{M}}$
$\therefore$ Initial value at $T_1$ $v_{rms1}=\sqrt{\frac{3R{T}_1}{M}}=\sqrt{\frac{3R\times 300}{M}}$ ... (1)
Final value at $T_2$ $v_{rms2}=\sqrt{\frac{3R{T}_2}{M}}=\sqrt{\frac{3R\times 1200}{M}}$.. (2)
[Molar mass is same as gas is same]
On dividing equation (2) by equation (1), we get
$\frac{v_{rms2}}{v_{rms1}}=\frac{\sqrt{\frac{3R\times 1200}{M}}}{\sqrt{\frac{3R\times 300}{M}}}$
$\frac{v_{rms2}}{v_{rms1}}=\sqrt{\frac{1200}{300}}$
$\Rightarrow v_{rms2}=2\times v_{rms1}$
The root mean square speed will be doubled when temperature changes from ${27}^{\circ }\text{C}$ to ${927}^{\circ }\text{C}$.