Question

As shown in the diagram, for a closed path $ABCA$

• A. the amount of work done by the system is zero
• B. the amount of work done by the system is $=-\frac{1}{2}(P_2-P_1)(V_2-V_1)$
• C. the amount of work done on the system is $=(P_2-P_1)(V_1-V_2)$
• D. the amount of work done by the system is $=\frac{1}{2}(P_2-P_1)(V_2-V_1)$

Answer 1

Answer: (B)

the amount of work done by the system is $=-\frac{1}{2}(P_2-P_1)(V_2-V_1)$

Given: P-V graph containing a closed path ABCA.

Solution:

For a closed cyclic path,work done is given by area of the closed path. If the cyclic process is in clockwise direction work done by the system is positive.

If the cyclic process is in anticlockwise direction work done by the system is negative.

Area of a triangle=$\frac{1}{2}\times base\times height$

So, area of the above triangle=$\frac{1}{2}\times (V_2-V_1)\times (P_2-P_1)$

As the cyclic process is in anticlockwise direction, work done by the system is negative.

$W=-\frac{1}{2}(P_2-P_1)(V_2-V_1)$

Hence the correct option is B.