JEE Main 25 Feb 2021 Shift 2 Solved Paper

Here, p1 and p2,T1 and T2,V1 and V2 are initial and final pressures, temperatures and volumes, respectively.
Given, ‌‌V2=2V1
pV1∕2=‌ constant ‌
From graph, γ= adiabatic constant =1∕2
⇒‌‌p1V11∕2=p2V21∕2⇒‌

p1

p2

=(‌

2V1

V1

)1∕2=21∕2
Also, ‌‌p1−γTγ= constant (for adiabatic process)
⇒‌‌p11−γT1γ=p21−γT2γ
⇒‌‌(‌

p1

p2

)1−γ=(‌

T2

T1

)γ⇒(‌

p1

p2

)‌

1−γ

γ

=‌

T2

T1

∴‌‌T2∕T1=(21∕2)‌

1−0.5

0.5

=√2