A vertical cylinder, closed at the bottom, holds a monatomic ideal gas under a light, frictionless…

Step 1. Final state (a). The spring is squeezed by $10\ \text{cm}$ and pushes down with $kx$, so

The volume grows by $Ax$ to $2.0\ \text{litres}$. From $PV/T$ fixed: $T_2 = 300\ \text{K} \times \dfrac{2.0 \times 10^{5} \times 2.0}{1.0 \times 10^{5} \times 1.0} = 1200\ \text{K}$.

Step 2. Work (b). The pressure grows in a straight line from $1.0 \times 10^{5}$ to $2.0 \times 10^{5}\ \text{Pa}$, so the area under the path is a trapezium:

Of this, $P_0 A x = 100\ \text{J}$ pushes back the air and $\tfrac12 kx^2 = 50\ \text{J}$ is stored in the spring.

Step 3. Heat (c). $\Delta U = \tfrac32(P_2V_2 - P_1V_1) = \tfrac32(400 - 100)\ \text{J} = 450\ \text{J}$, so $Q = \Delta U + W = 450 + 150 = 600\ \text{J}$.

Step 4. Heat capacity (d). The amount of gas is $n = \dfrac{P_1V_1}{RT_1} = 0.0401\ \text{mol}$, and $n\Delta T = 36.08\ \text{mol K}$. So