AP EAPCET 20-Aug-2021 Shift 1 Paper

According to given graph,As we know that, C=

R

γ−1

−

R

m−1

...(i)
where, CV is heat coefficient at constant volume,
R is universal gas constant =8.314JK−1 ,
γ is adiabatic constant and m is polytropic constant.
For path AB,pV2= constant
∴m=2 and γmonoatomic=

5

2

Put these values in Eq (i), we get
C=

R

5 3 −1

−

R

2−1

=

3R

2

−R=

R

2

For path BC, density is fixed
∴ Volume will remain unchanged.
Since, U=nCVΔT
where, U is internal energy
and CV (monoatomic gas) =

3

2

R.
Isochoric process, ΔV=0
But temperature is decreasing and CV=

3

2

R.
∴ Option (c) is incorrect.
For path CA (isothermal)
W=nRT‌ln‌

ρ2

ρ1

=nRT‌ln‌

4ρ0

ρ0

=nRT‌ln‌4
and U0=

3

2

‌nRT
∴ W=

2

3

U0(ln‌4)