Question

Calculate q for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure 1 bar to a final pressure 0.1 bar at constant temperature

of 273 K.

• A. -5227 J
• B. 5227 J
• C. 2240 J
• D. None of these

Answer 1

The correct option is B 5227 J

For isothermal reversible expansion

$W=-2.303nRTlog\frac{P_1}{P_2}$

$=-2.303\times 1\times 8.314\times 273log\frac{1.0}{0.1}=-5227.2J$

At constant temperature, for expansion $\Delta T=0$

Therefore , $\Delta E=0$ Also, $\Delta U=n\times C_v\times \Delta T=0$

And $\Delta U=q+W$ (for 1 mole)

$\therefore q=-W=5227.2J$