Question

One mole of an ideal gas is heated slowly so that it goes from the PV state (Pi, Vi) to (3Pi, 3Vi) in such a way that the pressure is directly proportional to the volume. (a) How much work is done on the gas in the process? (b) How is the temperature of the gas related to its volume during

this process?

Answer 1

At the expansion process the pressure is directly proportional to the volume

$$P=kV$$

The work done by definition

$$W=\int_{V_1}^{V_2}PdV=k\int_{V_1}^{V_2}VdV=\frac{kV_2^2}{2}-\frac{kV_1^2}{2}$$

$$W=\frac{k(3V_1)^2}{2}-\frac{kV_1^2}{2}=4kV_1^2=4P_1V_1$$

For the one mole of ideal gas the equation of state is given by

$$PV=RT$$

In our case we get

$$kV^2=RT$$

Hence

$$T\sim V^2$$