Kinetic Theory Of Gases Question 141
Question: Two thermally insulated vessels 1 and 2 are filled with air at temperature $ T _{1},,T _{2}; $ volumes $ V _{1},,V _{2} $ and pressures $ P _{1},,P _{2}, $ respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be
Options:
A) $ T _{1}=T _{2} $
B) $ (T _{1}+T _{2})/2 $
C) $ \frac{T _{1}T _{2}(P _{1}V _{1}+P _{2}V _{2})}{P _{1}V _{1}+P _{2}V _{2}T _{1}} $
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: C
Solution:
[c] The guiding principle in this problem is that the total number of moles of the system remain the same. $ \frac{P _{1}V _{1}}{RT _{1}}+\frac{P _{2}V _{2}}{RT _{2}}=\frac{P(V _{1}+V _{2})}{RT} $ or $ T=\frac{P(V _{1}+V _{2})T _{1}T _{2}}{P _{1}V _{1}T _{2}+P _{2}V _{2}T _{1}} $ By Boyle’s law, $ P(V _{1}+V _{2})=P _{1}V _{1}+P _{2}V _{2} $
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Therefore $ $ P=\frac{P _{1}V _{1}+P _{2}V _{2}}{V _{1}+V _{2}} $
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Therefore $ $ T=\frac{(P _{1}V _{1}+P _{2}V _{2})T _{1}T _{2}}{P _{1}V _{1}T _{2}+P _{2}V _{2}T _{1}} $
BETA
