One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is 4 cm$$^2$$. The gas is heated slowly to raise the temperature by 1.2 $$^\circ$$C during which the piston moves by 25 mm. The amount of heat supplied to the gas is __________ J. (Atmospheric pressure = 100 kPa, $$R = 8.3$$ J/mol. K) (Neglect mass of the piston)

Solution :

For one mole of diatomic gas with rotational modes only :

$$C_v = \frac{5R}{2}$$

Therefore,

$$C_p = C_v + R$$

$$= \frac{5R}{2} + R$$

$$= \frac{7R}{2}$$

Given :

$$n = 1$$

Temperature rise :

$$\Delta T = 1.2^\circ\text{C} = 1.2\text{ K}$$

Cross-sectional area :

$$A = 4\text{ cm}^2$$

$$= 4 \times 10^{-4}\text{ m}^2$$

Displacement of piston :

$$x = 25\text{ mm}$$

$$= 25 \times 10^{-3}\text{ m}$$

Change in volume :

$$\Delta V = Ax$$

$$= (4 \times 10^{-4})(25 \times 10^{-3})$$

$$= 10^{-5}\text{ m}^3$$

Atmospheric pressure :

$$P = 100\text{ kPa}$$

$$= 10^5\text{ Pa}$$

Work done :

$$W = P\Delta V$$

$$= 10^5 \times 10^{-5}$$

$$= 1\text{ J}$$

Heat supplied :

$$Q = nC_v\Delta T + W$$

$$= 1 \times \frac{5 \times 8.3}{2} \times 1.2 + 1$$

$$= 24.9 + 1$$

$$= 25.9\text{ J}$$

Final Answer :

$$25.9\text{ J}$$