IIT JEE Advanced 2011 Paper 1

For an adiabatic process, we have
T1V1γ−1 = T2V2γ−1
For monatomic gas γ = 5/3, and it is given that initial volume V1 = 5.6 L and temperature T1. Therefore,
T2 = T1(

V1

V2

)γ−1 = T1(

5.6

0.7

)(

5

3

)−1 = 8

2

3

×T1
For an adiabatic process, the work done is
W =

nR

γ−1

(T2−T1)
However, at STP, 1 mol of substance occupies 22.4 L.
Therefore,
n =

5.6

22.4

=

1

4

‌mol
W =

R(4T1−T1)

4[( 5 3 )−1]

=

9

8

RT1