CGPET 2021 Solved Paper

Given, number of mole, n=1
Initial temperature, Ti=400K
Since, final temperature, Tf=2Ti
∴ Tf=800K
Difference in temperature,
∆T=Tf−Ti=800K−400K=400K
Now, internal energy, ∆U=nCv.∆T
⇒ ∆U=1.(

3

2

R).400
[For monoatomic gas, Cv=

3

2

R]
∴ ∆U=600R.......(i)
According to the problem, process is given as
V2T=K ........(ii)
From the equation of ideal gas for 1 mole,
pV=RT⇒T=

pV

R

⇒ V2.(

pV

R

)=K[From‌Eq.(ii)]
⇒ V3p=K......(iii)
Standard equation of adiatomic process is
pVx=K
On comparing from Eq. (iii), we get
x=3
Since, the expression for molar heat capacity,
C=

R

γ−1

+

R

1−x

⇒ C=

R

( 5 2 −1)

+

R

1−3

(∵ For monoatomic gas, γ=

5

2

)
Solving above equation, we get C=R and heat required,
∆Q=n.C.∆T=1×R×400=400R .......(iv)
Thus, work done, ∆W=∆Q−∆U
From Eqs. (i) and (iv), we get
∆W=400R−600R=−200R