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(C) The work done by a system during a process is equal to the area under the $PV$ graph.In this case,the area under the line represents a trapezium.T

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$12 	imes 10^5 	ext{ J}$

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(C) The work done by a system during a process is equal to the area under the $PV$ graph.In this case,the area under the line represents a trapezium.The parallel sides of the trapezium are the pressure values $P_1 = 1 	imes 10^5 	ext{ N/m}^2$ and $P_2 = 5 	imes 10^5 	ext{ N/m}^2$.The height of the trapezium is the change in volume,$\Delta V = V_2 - V_1 = 5 	ext{ m}^3 - 1 	ext{ m}^3 = 4 	ext{ m}^3$.The area of a trapezium is given by $	ext{Area} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$.$	ext{Work done} = \frac{1}{2} 	imes (1 	imes 10^5 + 5 	imes 10^5) 	imes (5 - 1)$$	ext{Work done} = \frac{1}{2} 	imes (6 	imes 10^5) 	imes 4$$	ext{Work done} = 3 	imes 10^5 	imes 4 = 12 	imes 10^5 	ext{ J}$.

$A$ system changes from the state $(P_1, V_1)$ to $(P_2, V_2)$ as shown in the figure. What is the work done by the system?

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