JEE Main On 08 April 2018 Question 29

Question: Two moles of an ideal monoatomic gas occupies a volume $ V $ at $ 27{}^\circ C $ . The gas expands adiabatically to a volume $ \text{2 V} $ . Calculate the final temperature of the gas and change in its internal energy. [JEE Main Online 08-04-2018]

Options:

A) $ \text{(A) 189 K}\text{(B) -2}\text{.7 kj} $

B) $ \text{(A) 195 K}\text{(B) 2}\text{.7 kj} $

C) $ \text{(A) 189 K}\text{(B) 2}\text{.7 kj} $

D) $ \text{(A) 195 K}\text{(B) -2}\text{.7 kj} $

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Answer:

Correct Answer: A

Solution:

  • $ n=2,T _1=27{}^\circ C=300K,V _{i}=V,V _{f}=2V $ In adiabatic condition $ T _1V _1^{\gamma -1}=T _2V _2^{\gamma -1} $ $ 300\times {V^{(5/3-1)}}=T _2{{(2V)}^{5/3-1}} $
    $ \Rightarrow T _2\approx 189 $
    $ \therefore \Delta U=nC _{V}\Delta T $ As temp decreases so $ \Delta U $ is - ve $ \Delta U=2\times ( \frac{R}{\gamma -1} )\Delta T=-2.7kJ $