TS EAMCET 19-July-2022 Shift 1 Solved Paper

In given process
Pressure ∝ Volume ⇒p=kV
work done in process,
∆W‌=

3V0

∫

V0

ρdV=

3V0

∫

V0

kV⋅dV
or ∆W=4p0V0......(i)
As gas is monoatomic, CV=‌

3

2

R
Change in internal energy of gas,
∆U=nCv(Tf−Ti).....(ii)
Now, from pV=nRT;
Ti=‌

p0V0

nR

and Tf=‌

9p0V0

nR

So, from Eq. (ii) ∆U=n⋅‌

3

2

R⋅‌

8p0V0

nR

=12p0V0
By ∆Q=∆U+∆W, we have
Heat absorbed ∆Q=12p0V0+4p0V0=16p0V0