Question

An ideal gas initially at temperature, pressure and volume, $ 27^{\circ}C $ , $ 1.00 $ bar and $ 10\, L $ , respectively is heated at constant volume until pressure is $ 10.0 $ bar, it then undergoes a reversible isothermal expansion until pressure is $ 1.00 $ bar, what is the total work $ W $ , during this process?

• A. $ -23.02 \times 10^{3} \, J $
• B. $ -14.0 \times 10^{3}\, J $
• C. $ 14.0 \times 10^{3}\,J $
• D. Zero

Answer 1

Answer: (A)

Given, $p_{i}=1$ bar
$V_{i}=10\,L$
$\because$ For ideal gas,
$PV=nRT$
$\therefore n=\frac{PV}{RT}=\frac{1\times10}{RT}$
$\because$ Work done $=-p\Delta V$
$\therefore $ Work done at constant volume
i.e when $\Delta V=0$
$W=0 $
Work done in isothermal reversible process
Now, $W=-2.303\, nRT \, log \frac{p_{i}}{p_{t}}$
$p_{i}=10 \,bar\, n=\frac{10}{RT}$
$\therefore W=2.303\times\frac{10}{RT} log \frac{10}{1}$
$W=-2303\times10=-23.02 \,atm \,L $
$\because 1\,atm \, L=10^{5}N-m^{2}$ and $1\,N-m^{2}=1\,J $
$\therefore W=-23.02\times10^{5}J$