Initial pressure and volume of a monoatomic ideal gas are $$P$$ and $$V$$. The change in internal energy of this gas in adiabatic expansion to volume $$V_{final} = 27V$$ is __________ J.

Solution :

For a monoatomic ideal gas,

$$\gamma = \frac{5}{3}$$

In adiabatic process :

$$PV^\gamma = \text{constant}$$

Given :

Initial pressure and volume :

$$P_i = P,\quad V_i = V$$

Final volume :

$$V_f = 27V$$

Therefore,

$$P_iV_i^\gamma = P_fV_f^\gamma$$

$$PV^{5/3} = P_f(27V)^{5/3}$$

Since,

$$27^{5/3} = (3^3)^{5/3} = 3^5 = 243$$

$$PV^{5/3} = 243P_fV^{5/3}$$

$$P_f = \frac{P}{243}$$

Internal energy of monoatomic gas :

$$U = \frac{3}{2}PV$$

Initial internal energy :

$$U_i = \frac{3}{2}PV$$

Final internal energy :

$$U_f = \frac{3}{2}P_fV_f$$

$$= \frac{3}{2}\times \frac{P}{243}\times 27V$$

$$= \frac{3}{2}\times \frac{PV}{9}$$

$$= \frac{PV}{6}$$

Change in internal energy :

$$\Delta U = U_f-U_i$$

$$= \frac{PV}{6}-\frac{3PV}{2}$$

$$= \frac{PV-9PV}{6}$$

$$= -\frac{8PV}{6}$$

$$= -\frac{4PV}{3}$$

Final Answer :

$$-\frac{4PV}{3}\text{ J}$$