The pressure of a gas changes linearly with volume from $$A$$ to $$B$$ as shown in figure. If no heat is supplied to or extracted from the gas then change in the internal energy of the gas will be

Solution :

From first law of thermodynamics :

$$\Delta Q = \Delta U + W$$

Given, no heat is supplied or extracted :

$$\Delta Q = 0$$

Therefore,

$$\Delta U = -W$$

From graph, process occurs from A to B.

Coordinates :

At \(A\),

$$P_A = 10\text{ kPa}$$

$$V_A = 200\text{ cc}$$

At \(B\),

$$P_B = 50\text{ kPa}$$

$$V_B = 50\text{ cc}$$

Work done by gas is area under \(PV\) graph :

$$W = \frac{(P_A+P_B)}{2}(V_B-V_A)$$

$$= \frac{(10+50)}{2}(50-200)$$

$$= 30 \times (-150)$$

$$= -4500\text{ kPa-cc}$$

Now,

$$1\text{ kPa-cc} = 10^{-3}\text{ J}$$

Therefore,

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

$$= -4.5\text{ J}$$

Hence,

$$\Delta U = -(-4.5)$$

$$= 4.5\text{ J}$$

Final Answer :

$$4.5\text{ J}$$