Answer:-

Given:

• Initial volume (\(V_1\)) = 500 cm³
• External pressure (\(P_{\text{ext}}\)) = 1.2 × 10⁵ Pa
• Work (\(W\)) = 36.0 J

To find: Final volume (\(V_2\))

Formula: \(W = -P_{\text{ext}} \Delta V = -P_{\text{ext}} (V_2 - V_1)\)

Calculation:

• Initial volume (\(V_1\)) = 500 cm³ = 0.5 dm³
• External pressure (\(P_{\text{ext}}\)) = 1.2 × 10⁵ Pa = 1.2 bar
• Work (\(W\)) = 36.0 J = 0.360 dm³ bar (using \(1 \, \text{dm}^3 \, \text{bar} = 100 \, \text{J}\))

Now, from the formula:

\[W = -P_{\text{ext}} \Delta V = -P_{\text{ext}} (V_2 - V_1)\]

Therefore:

\[\frac{0.360 \, \text{dm}^3 \, \text{bar}}{1.2 \, \text{bar}} = -(V_2 - 0.5 \, \text{dm}^3)\]

So:

\[0.3 \, \text{dm}^3 = -V_2 + 0.5 \, \text{dm}^3\]

Thus:

\[V_2 = 0.2 \, \text{dm}^3 = 200 \, \text{cm}^3\]

The final volume (\(V_2\)) is 200 cm³.