AP EAPCET 08-Jul-2022 Shift 2 Paper

Work done by gas in adiabatic process,
∆W=+83J
As in adiabatic process, ∆Q=0,
By first law of thermodynamics,
∆Q=∆U+∆W
⇒0=∆U+∆W
⇒∆U=−∆W
so change in internal energy of gas,
∆U=−83J
Also, ∆U=nCV∆T=83
Here, n=1 mole
⇒ specific heat at constant volume
CV=Cp−R
⇒CV=

11

10

R−R=

1

10

R
So, −83=1×

1

10

R×(Tf−125)
⇒Tf−125=

−830

8.3

=−100
or Tf=125−100=25∘C