Question

Figure shows three paths through which a gas can be taken from the state A to the state B. Calculate the work done by the gas in each of the three paths.

Answer 1

Work done during any process, W = P ∆ V

If both pressure and volume are changing during a process, then work done can be found out by finding the area under the PV diagram.

In path ACB, for line AC :-

Since initial volume is equal to final volume,

∆ V = 0

⇒ W_AC = P ∆ V = 0

For line BC :-

P = 30 × 10³ pa

W_ACB = W_AC + W_BC = 0 + P∆V

= 30 × 10³ × (25 − 10) × 10⁻⁶

= 0.45 J

For path AB:-

Since both pressure and volume are changing, we use the mean pressure to find the work done.

Mean pressure, P = \[\frac{1}{2} \times (30 + 10) \times  {10}^3\]

W_AB = \[\frac{1}{2}\]× (10 + 30) × 10³ × 15 × 10⁻⁶

= \[\frac{1}{2}\] × 40 × 15 × 10⁻³ = 0.30 J

Initial volume in path ADB, along line DB is the same as final volume. Thus, work done along this line is zero.

Along line AD, P = 10 kPa

W = W_AD + W_DB

= 10 × 10³ (25 − 10) × 10⁻⁶ + 0

= 10 × 15 × 10⁻³ = 0.15 J