TS EAMCET 18-July-2022 Shift 1 Solved Paper

Given, W=100J
From ideal gas equation,
p∆V‌=nR∆T‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
W‌=p∆V
∴ Eq. (i) becomes,
W‌=nR∆T
∆T‌=‌

W

nR

=‌

100

nR

‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
Now, we know the relation for change in internal energy,
∆U=nCV∆T
⇒‌‌∆U=n‌

R

γ−1

∆T‌‌... (iii) (∵CV=‌

R

γ−1

) and since it is a monoatomic gas,
∴‌‌γ=‌

5

3

From Eqs. (ii) and (iii)
⇒‌‌∆U=150J
⇒‌‌∆U=‌

nR∆T

5 3 −1

=‌

3nR

2

×‌

100

nR

=150J
Now, the heat given to the gas in the process is
∆Q=∆U+W
⇒‌‌∆Q=150+100=250J
⇒‌‌∆Q=250J