Kinetic Theory Of Gases Question 155
Question: DIRECTION: Read the passage given below and answer the questions that follows:
A diathermic piston divides adiabatic cylinder volume $ V _{0} $ into two equal parts as shown in the figure. Both parts contain ideal monoatomic gases. The initial pressure and temperature of gas in left compartment are $ P _{0} $ and $ T _{0} $ while that in right compartment are $ 2P _{0} $ and $ 2T _{0} $ . Initially the piston is kept fixed and the system is allowed to acquire a state of thermal equilibrium. If the pin which was keeping the piston fixed is removed and the piston is allowed to slide slowly such that a state of mechanical equilibrium is achieved. The volume of the left compartment when the piston is in equilibrium is
Options:
A) $ \frac{3}{4}V _{0} $
B) $ \frac{V _{0}}{4} $
C) $ \frac{V _{0}}{2} $
D) $ \frac{2}{3}V _{0} $
Show Answer
Answer:
Correct Answer: C
Solution:
Let AV be the change in volume in any compartment then $ n _{1}=\frac{P _{0}V _{0}}{2RT _{0}}=\frac{P _{f}( \frac{V _{0}}{2}-\Delta V )}{RT _{f}} $ and $ n _{2}=\frac{2P _{0}V _{0}}{2RT _{0}}=\frac{P _{f}( \frac{V _{0}}{2}+\Delta V )}{RT _{f}}\Rightarrow \Delta V=\frac{P _{f}V _{0}}{2RT _{f}}-\frac{V _{0}}{2} $
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