Question

An ideal gas initially at temperature, pressure and volume, $27^{\circ }C$ , $1.00$ bar and $10L$ , 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
$V_i=10L$
$\because$ For ideal gas,
$PV=nRT$
$\therefore n=\frac{PV}{RT}=\frac{1\times 10}{RT}$
$\because$ 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, $W=-2.303nRTlog\frac{p_i}{p_t}$
$p_i=10barn=\frac{10}{RT}$
$\therefore W=2.303\times \frac{10}{RT}log\frac{10}{1}$
$W=-2303\times 10=-23.02atmL$
$\because 1atmL=10^{5}N-m^2$ and $1N-m^2=1J$
$\therefore W=-23.02\times 10^{5}J$