Kinetic Theory Of Gases Question 170
Question: A partition divides a container having insulated walls into two compartments I and II. The same gas fills the two compartments whose initial parameters are given. The partition is a conducting wall which can move freely without friction. Which of the following statement is correct, with reference to the final equilibrium position?
Options:
A) The Pressure in the two compartments are unequal.
B) Volume of compartment I is $ \frac{2V}{5} $
C) Volume of compartment II is $ \frac{12,V}{5} $
D) Final pressure in compartment I is $ \frac{4P}{3} $
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Answer:
Correct Answer: C
Solution:
[c] In the equilibrium position the net force on the partion will be zero. Hence pressure on both sides are same. Initially, PV = nRT $ n _{1}=\frac{P _{1}V _{1}}{RT _{1}}=\frac{PV}{RT} $ & $ ,n _{2}=\frac{( 2P )( 2V )}{RT}=4\frac{PV}{RT}\Rightarrow n _{2}=4n _{1} $ Moles remains conserved. Finally, pressure becomes equal in both parts. Using, $ P _{1}V _{1}=n _{1}RT _{1} $ $ P _{2}V _{2}=n _{2}RT _{2} $
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Therefore $ $ P _{1}=P _{2}\And T _{1}=T _{2} $
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Therefore $ $ \frac{V _{1}}{V _{2}}=\frac{n _{1}}{n _{2}}=\frac{1}{4}\Rightarrow ,V _{2}=4V _{1} $ Also $ V _{1}+V _{2}=3V $
$ \Rightarrow $ $ V _{1}+4V _{1}=3V $ And $ V _{2}=\frac{12}{5}V $ In compartment (I): $ P _{1}‘V _{1}=n _{1}RT _{1} $ $ P _{1}( \frac{3V}{5} )=( \frac{PV}{RT} )RT $ $ P _{1}’=\frac{5PV}{3V}=\frac{5}{3}P $
BETA
