Question

An ideal gas is taken from state $A$ to the state $B$, as shown in the $P-V$ diagram. The work done in the process is

• A. $(P_A-P_B)\times (V_B-V_A)$
• B. $\frac{1}{2}(P_B-P_A)\times (V_B+V_A)$
• C. $\frac{1}{2}(P_B-P_A)\times (V_B-V_A)$
• D. $\frac{1}{2}(P_B+P_A)\times (V_B-V_A)$

Answer 1

The correct option is D $\frac{1}{2}(P_B+P_A)\times (V_B-V_A)$

We know that,
Work done from state $A$ to the state $B=$ Area under $P-V$ graph from $A$ to $B$
From the figure, we can say that
Work done = Area of trapezium under straight line $AB$
$W=\frac{1}{2}(P_A+P_B)\times (V_B-V_A)$
Hence, option (d) is the correct answer.