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) -2.7 kJ} $
B) $ \text{(A) 195 K}\text{(B) 2.7 kJ} $
C) $ \text{(A) 189 K}\text{(B) 2.7 kJ} $
D) $ \text{(A) 195 K}\text{(B) -2.7 kJ} $
Show Answer
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 temperature decreases so $ \Delta U $ is negative
$ \Delta U=2\times ( \frac{R}{\gamma -1} )\Delta T=-2.7kJ $
BETA
