TS EAMCET 5 May 2018 Shift 2 Solved Paper

According to adiabatic process,
TVγ−1= constant
Form the given conditions,
V1=V and V2=(

V

8

) and The value of adiabatic index for monoatomic gas is γ=

5

3

.
So,
T1V

5

3

−1=T2(

V

8

)

5

3

−1
=

T2

T1

=

V 2 3

( V 8 ) 2 3

=8

2

3

=4
So, temperature of gas increases by 4 times. As mean KE of gas molecule is proportional to its temperature, then KE also increases by 4 times.