A gas undergoes a process where its pressure varies linearly with...

The work done by the gas in a process where pressure varies linearly with volume is given by the area under the P-V graph, which forms a trapezium.

Initial pressure $P_1 = 10 \text{ kPa} = 10 \times 10^3 \text{ Pa}$

Final pressure $P_2 = 50 \text{ kPa} = 50 \times 10^3 \text{ Pa}$

Initial volume $V_1 = 200 \text{ cc} = 200 \times 10^{-6} \text{ m}^3$

Final volume $V_2 = 50 \text{ cc} = 50 \times 10^{-6} \text{ m}^3$

The work done $W$ is calculated as:

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

$W = \dfrac{1}{2}(10 \times 10^3 + 50 \times 10^3)(50 \times 10^{-6} - 200 \times 10^{-6})$

$W = \dfrac{1}{2}(60 \times 10^3)(-150 \times 10^{-6})$

$W = 30 \times 10^3 \times (-150 \times 10^{-6})$

$W = -4500 \times 10^{-3} \text{ J}$

$W = -4.5 \text{ J}$