Question

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

• A. $q =2.303 nRT \log \frac{ V _{2}}{ V _{1}}$
• B. $q =-23 \cdot 3032 n RT \log \frac{ V _{2}}{ V _{1}}$
• C. $q=-2.303 n R T \log \frac{v_{1}}{v_{2}}$
• D. $q=-P_{\exp } n R T \log \frac{v_{1}}{v_{2}}$

Answer 1

Answer: (A)

For an isothermal reversible process, $\Delta T =0$
$
\therefore \Delta U = nC _{ V } \Delta T =0
$
$
\begin{array}{l}
\text { Also, } \Delta U = q + w =0 \\
\therefore q =- W
\end{array}
$
For an isothermal reversible expansion, $w=-n R T \ln \left(\frac{V_{2}}{V_{1}}\right)=$
$-2.303 nRT \log \left(\frac{ V _{2}}{ V _{1}}\right)$ $\therefore q =- w =2.303 nRT \log \left(\frac{ V _{2}}{ V _{1}}\right)$