Question
A system is taken to its final state from initial state in hypothetical paths as shown figure. Calculate the work done in each case.
✓
Answer
Data: $P_A=P_B=6 \times 10^5 \mathrm{~Pa}, \mathrm{P}_{\mathrm{C}}=\mathrm{P}_{\mathrm{D}}=2 \times 10^5 \mathrm{~Pa} \mathrm{~V}_{\mathrm{A}}=$ $V_D=2 L_{,} V_B=V_C 6 L, 1 \mathrm{~L}=10^{-3} \mathrm{~m}^3$ (i) The work done along the path $A \rightarrow B$ (isobaric process), $W_{A B}=P_A\left(V_B-V_A\right)=\left(6 \times 10^5 \mathrm{~Pa}\right)(6-2)\left(10^{-3}\right.$ $\left.m^3\right)=2.4 \times 10^3 \mathrm{~J}$ (ii) $\mathrm{W}_{\mathrm{BC}}=$ zero as the process is isochoric $\mathrm{V}=$ constant). (iii) The work done along the path $\mathrm{C} \rightarrow \mathrm{D}$ (isobaric process), $W_{C D}=P_C\left(V_D-V_C\right)$ $=\left(2 \times 10^5 \mathrm{~Pa}\right)(2-6)\left(10^{-3} \mathrm{~m}^3\right)=-8 \times 10^2 J$ (iv) $\mathrm{W}_{\mathrm{DA}}=$ zero as $\mathrm{V}=$ constant.
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