Question

A thermodynamic system goes from states (i) $P_{1}$, V to $2P_{1}$, V (ii) P, V to P, 2V. Then work done in the two cases is

• A. Zero, Zero
• B. Zero, $PV_{1}$
• C. $PV_{1}$, Zero
• D. $PV_{1},\mspace{6mu} P_{1}V_{1}$

Answer 1

Answer: (B)

Zero, $PV_{1}$

Answer 2

(i) Case $\rightarrow$Volume = constant $\Rightarrow \int_{}^{}{PdV} = 0$

(ii) Case$\rightarrow$P = constant $\Rightarrow \int_{V_{1}}^{2V_{1}}{PdV} = P\int_{V_{1}}^{2V_{1}}{dV = PV_{1}}$