Thermodynamics Question 237
Question: An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $ V _{1} $ and contains ideal gas at pressure $ P _{1} $ , and temperature $ T _{1} $ The other chamber has volume $ V _{2} $ and contains ideal gas at pressure $ P _{2} $ and temperature $ T _{2} $ If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be
Options:
A) $ \frac{T _{1}T _{2}( P _{1}V _{1}+P _{2}V _{2} )}{P _{1}V _{1}+P _{2}V _{2}} $
B) $ \frac{P _{1}V _{1}T _{1}+P _{2}V _{2}T _{2}}{P _{1}V _{1}+P _{2}V _{2}} $
C) $ \frac{P _{1}V _{1}T _{2}+P _{2}V _{2}T _{1}}{P _{1}V _{1}+P _{2}V _{2}} $
D) $ \frac{T _{1}T _{2}( P _{1}V _{1}+P _{2}V _{2} )}{P _{1}V _{1}T _{1}+P _{2}V _{2}T _{2}} $
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Answer:
Correct Answer: A
Solution:
[a] Here $ Q=0 $ and $ W=0. $
Therefore from first law of thermodynamics $ \Delta U=Q+W=0 $
$ \therefore $ Internal energy of the system with partition = Internal energy of the system without partition. $ n _{1}C _{v}T _{1}+n _{2}C _{v}T _{2}=( n _{1}+n _{2} )C _{v}T $
$ \therefore T=\frac{n _{1}T _{1}+n _{2}T _{2}}{n _{1}+n _{2}} $
$ \text{But }n _{1}=\frac{P _{1}V _{1}}{RT _{1}}\text{ and }n _{2}=\frac{P _{2}V _{2}}{RT _{2}} $
$ \therefore T=\frac{\frac{P _{1}V _{1}}{RT _{1}}\times T _{1}+\frac{P _{2}V _{2}}{RT _{2}}\times T _{2}}{\frac{P _{1}V _{1}}{RT _{1}}+\frac{P _{2}V _{2}}{RT _{2}}}=\frac{T _{1}T _{2}( P _{1}V _{1}+P _{2}V _{2} )}{P _{1}V _{1}T _{2}+P _{2}V _{2}T _{1}} $
