Kinetic Theory Of Gases Question 169
Question: A box contains N molecules of a perfect gas at temperature $ T _{1} $ and pressure $ P _{1} $ . The number of molecules in the box is doubled keeping the total kinetic energy of the gas same as before. If the new pressure is $ P _{2} $ and temperature $ T _{2}, $ then
Options:
A) $ P _{2}=P,T _{2}=T _{1} $
B) $ P _{2}=P _{1},,T _{2}=\frac{T _{1}}{2} $
C) $ P _{2}=2P _{1},,T _{2}=T _{1} $
D) $ P _{2}=2P _{1},,T _{2}=\frac{T _{1}}{2} $
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Answer:
Correct Answer: B
Solution:
[b] Kinetic energy of N molecule of gas, $ E=\frac{3}{2}NkT $ Initially, $ E _{1}=\frac{3}{2}N _{1}kT _{1} $ and finally, $ E _{2}=\frac{3}{2}N _{2}kT _{2} $ But according to $ E _{1}=R _{2} $ and $ N _{2}=2N _{1}, $ $ \frac{3}{2}N _{1}kT _{1}=\frac{3}{2},(2N _{1})kT _{2}\Rightarrow ,T _{2}=\frac{T _{1}}{2} $ Since the kinetic energy constant, $ \frac{3}{2}N _{1}kT _{1}=\frac{3}{2}N _{2}kT _{2} $
$ \Rightarrow $ $ N _{1}T _{1}=N _{2}T _{2} $
$ \
Therefore $ $ NT= $ Constant From ideal gas equation of N molecules, $ PV=NkT $
$ \Rightarrow $ $ P _{1}V _{1}=P _{2}V _{2}\Rightarrow ,P _{1}=P _{2} $ [as $ V _{1}=V _{2} $ and $ NT= $ constant]
BETA
