Question

In an isothermal irreversible expansion of an ideal gas:

• A. $\Delta U\ne 0$
• B. $-W=q=nRT[1-\frac{P_2}{P_1}]$
• C. $\Delta H\ne 0$
• D. All are correct

Answer 1

The correct option is B $-W=q=nRT[1-\frac{P_2}{P_1}]$

The internal energy of an ideal gas is directly proportional to the temperature of the gas.
$\Delta E=n{C}_v\Delta T=q+W$
In an isothermal process, there is no change in temperature so $\Delta T$ will be equal to zero. Hence, change in internal energy, $\Delta U=0,q=-W=P_2\Delta V=nRT[1-\frac{P_2}{P_1}]$