Question

One mole of an ideal gas at 300K is heated at constant volume $\left(V_1\right)$ until its temperature is doubled, then it is expanded isothermally till it reaches the original the pressure. Finally, the gas is cooled at the constant pressure till system reached to the half of original volume $\left(\frac{V_1}{2}\right)$. Determine the total work done $\left(w\right)$ in calories. [Use $ln2=0.70,0R=2cal{K}^{-1}mo{l}^{-1}$]

Answer 1

$1^{st}$ case

For constant volume work done is zero, so in first case work done is $=0$

as $w=-P\times \Delta V$ here $\Delta V=0$ so $w=0$

$2^{nd}$ case

Applying ideal gas law to determine the pressure,

$PV=nRT$ or $P_1=\frac{2\times 300}{V_1}$

now the temp is doubled at cons volume so $P_2$,

$P_2=\frac{P_1\times T_2}{T_1}=\frac{1200}{V_1}$

$W_2=-2.303nRTlog\frac{P_1}{P_2}=-2.303\times 1\times 2\times 300\times \frac{1}{2}=+414.54$

$3^{rd}$ case

$W_3=-P_2\times \Delta V$

$W_3=-(\frac{1200}{V_1}-\frac{V_1}{2})=600cal$

So total work done$=w_1+w_2+w_3=0+414.54+600cal=1.014Kcal$