The temperature of a gas is raised from 27^ C to 927^ C . The root mean square speed

Correct answer: B. gets doubled Given, aligned T_1 & =27^ C=300 K T_2 & =927^ C=1200 K aligned v_ rms = ((3 k T)/(m)) aligned & v_ rms T & (v_ rms )_1 (v_ rms )_2 = (T_1)/(T_2) & (v_ rms )_1 (v_ rms )_2 = (300)/(1200) =(1)/(2) & (v_ rms )_2=2(v_ rms )_1 & aligned

JIPMER 2016

Exam: JIPMER
Year: 2016
Subject: Physics
Chapter: Kinetic Theory of Gases

JIPMER 2016 · Physics · Kinetic Theory of Gases

The temperature of a gas is raised from 27^ C to 927^ C . The root...

The temperature of a gas is raised from

27^{\circ} \mathrm{C}

927^{\circ} \mathrm{C}

. The root mean square speed

Options

Agets halved
Bgets doubled
Cis $\sqrt{\left(\frac{927}{27}\right)}$ times the earlier value
Dremains the same

Correct answer

B. gets doubled

Step-by-step solution

Given,

\begin{aligned} T_1 & =27^{\circ} \mathrm{C}=300 \mathrm{ K} \\ T_2 & =927^{\circ} \mathrm{C}=1200 \mathrm{ K}\end{aligned}

v_{\mathrm{rms}}=\sqrt{\left(\frac{3 k T}{m}\right)}

\begin{aligned} & \Rightarrow \quad v_{\mathrm{rms}} \propto \sqrt{T} \\ & \therefore \quad \frac{\left(v_{\text{rms }}\right)_1}{\left(v_{\text{rms }}\right)_2}=\sqrt{\frac{T_1}{T_2}} \\ & \frac{\left(v_{\text{rms }}\right)_1}{\left(v_{\text{rms }}\right)_2}=\sqrt{\frac{300}{1200}}=\frac{1}{2} \\ & \Rightarrow \quad\left(v_{\mathrm{rms}}\right)_2=2\left(v_{\mathrm{rms}}\right)_1 \\ & \end{aligned}

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