Kinetic Theory Of Gases Question 4
Question: The molecules of a given mass of a gas have root mean square speeds of $ 100,m{{s}^{-1}} $ at $ 27{}^\circ C $ and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at $ 127{}^\circ C $ and 2.0 atmospheric pressure?
Options:
A) $ \frac{150}{\sqrt{3}}m/s $
B) $ \frac{125}{\sqrt{3}}m/s $
C) $ \frac{200}{\sqrt{3}}m/s $
D) $ 100\sqrt{3}m/s $
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Answer:
Correct Answer: C
Solution:
[c] We know that for a given mass of a gas $ v _{rms}=\sqrt{\frac{3RT}{M}} $ Where R is gas constant, T is temperature in kelvin and M is molar mass of the gas. Clearly for a given gas, $ v _{rms}\propto \sqrt{T} $ , as R, M are constants. Hence,
$ \frac{(v_rms)_1}{(v_rms)_2}$
=$\sqrt{\frac{T _{1}}{T _{2}}}…(i) $
Given, $ {{(v _{rms})} _{1}}=100m/s $
$ T _{1}=27{}^\circ C=27+273=300K $
$ T _{2}=127{}^\circ C=127+273=400K $
$ \
Therefore $ Form Eq. (i) $ \frac{100}{{{(v _{rms})} _{2}}}=\sqrt{\frac{300}{400}}=\frac{\sqrt{3}}{2} $
$ \Rightarrow {{(v _{rms})} _{2}}=\frac{2\times 100}{\sqrt{3}}=\frac{200}{\sqrt{3}}m/s $
BETA
