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

Given initial state:
$P_1 = 10 \text{ kPa} = 10 \times 10^3 \text{ Pa}$
$V_1 = 200 \text{ cc} = 200 \times 10^{-6} \text{ m}^3$

Final state:
$P_2 = 50 \text{ kPa} = 50 \times 10^3 \text{ Pa}$
$V_2 = 50 \text{ cc} = 50 \times 10^{-6} \text{ m}^3$

Since the pressure varies linearly with volume, the process on the P-V diagram is a straight line. The work done by the gas is the area under the P-V graph, which forms a trapezium:
$W = \dfrac{1}{2}(P_1 + P_2)(V_2 - V_1)$

Substituting the values:
$W = \dfrac{1}{2}(10 \times 10^3 + 50 \times 10^3)(50 \times 10^{-6} - 200 \times 10^{-6})$
$W = \dfrac{1}{2}(60 \times 10^3)(-150 \times 10^{-6})$
$W = 30 \times 10^3 \times (-150 \times 10^{-6}) = -4.5 \text{ J}$

According to the first law of thermodynamics:
$\Delta Q = \Delta U + W$

Since no heat is added to or extracted from the gas, $\Delta Q = 0$.
$0 = \Delta U + (-4.5 \text{ J})$
$\Delta U = 4.5 \text{ J}$