TS EAMCET 6-Aug-2021 Shift 1 Question Paper

The process in which there is no heat exchange of gas, it is called adiabatic process.
In this process, TVγ−1=k [constant]
Now, T2V2(γ−1)=k2= constant . . . (i)
Given that, T2Vα= constant . . . (ii)
Comparing Eqs. (i) and (ii), we get
α=2(γ−1)
Using γ=‌

4

3

, for polyatomic gas
α=2(‌

4

3

−1)=‌

2

3