Question

For an isothermal reversible expansion process, the value of q can be calculated by the expression :

• A. q = 2.303nRT log $\frac{V_2}{V_1}$
• B. q = -23.3032nRT log $\frac{V_2}{V_1}$
• C. q = -2.303nRT log $\frac{V_1}{V_2}$
• D. q = - $P_{exp}$nRT log $\frac{V_1}{V_2}$

Answer 1

The correct option is A q = 2.303nRT log $\frac{V_2}{V_1}$

For an isothermal reversible process, $\Delta T=0$

$\therefore \Delta U=n{C}_V\Delta T=0$

Also, $\Delta U=q+w=0$

$\therefore q=-w$

For an isothermal reversible expansion, $w=-nRTln\left(\frac{V_2}{V_1}\right)=-2.303nRTlog\left(\frac{V_2}{V_1}\right)$

$\therefore q=-w=2.303nRTlog\left(\frac{V_2}{V_1}\right)$