Question

$16g$ of $O_2$ at ${28}^{o}C$ is compressed to half of its initial volume under reversible isothermal conditions. If this gas behaves ideally, the work done and change in internal energy in this process are:

• A. $\Delta U=0$ and $w=-1247.1\ln 2J$
• B. $\Delta U=0$ and $w=+1247.1\ln 2J$
• C. $w=\Delta U=+1247.1\ln 2J$
• D. $\Delta U=-1247.1\ln 2J$ and $w=+1247.1\ln 2J$

Answer 1

Answer: (A)

$\Delta U=0$ and $w=-1247.1\ln 2J$

Internal energy depends on temperature. As in isothermal process, T is constant, so change in internal energy is zero.

$n=\frac{32}{16}=\frac{1}{2}$

$T={28}^{\circ }C=301K$

$W=-2.303nRT\log \left[\frac{V_2}{V_1}\right]$

$=nRTln\left[\frac{V_2}{V_1}\right]$

$=\frac{1}{2}\times 8.314\times 301ln\left[\frac{1}{2}\right]$

$=1251ln[2{]}^{-1}$

$=-1251ln\left[2\right]$