Question

One mole of a perfect monoatomic gas is put through a cycle consisting of the following three reversible steps:
(CA): Isothermal compression from 2 atm and 10 litres to 20 atm and 1 litre.
(AB): Isobaric expansion to return the gas to the original volume of 10 litres with T going from $T_{1}to{T}_2$.
(BC): Cooling at constant volume to bring the gas to the original pressure and temperature.
The steps are shown schematically in the figure given above
(a) Calculate $T_{1}to{T}_2$.
(b) Calculate $\Delta U$, q and w in calories, for each step and for the cycle.

Answer 1

We know,
Path CA-Isothermal compression
Path AB-Isobaric expansion
Path BC=Isochoric change
Let $V_i$ and $V_f$ are initial volume and final volume at respective points.
For temperature $T_1$ (For C):PV$={nRT}_1$
$2\times 10=1\times 0.0821\times T_1$
$\therefore T_1=243.60K$
For temperature $T_2\left(ForCandB\right):\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$\frac{2\times 10}{T_1}=\frac{20\times 10}{T_2}$
$\therefore \frac{T_2}{T_1}=10$
$\therefore T_2=243.60\times 10=2436.0K$
$PathCA:$ $\Delta U=0$ for isothermal compression; Also $q=w$
$w=+2.303{nRT}_1\log \frac{V_i}{V_f}$
$=2.303\times 1\times 2\times 243.6\times \log \frac{10}{1}$
$=+1122.02cal$
$PathAB:w=-P(V_f-V_i)$
$w=-20\times (10-1)=-180litre-atm$
$=\frac{-180\times 2}{0.0821}=-4384.9cal$
$PathBC:w=-P(V_f-V_i)=0$
$(\therefore V_f-V_i=0;sincevolumeisconstant)$
For monoatomic gas heat change at constant volume$=q_r=\Delta U$
Thus for path BC $q_r=C_v\times n\times \Delta T=\Delta U$
$q_r=\frac{3}{2}R\times 1\times (2436-243.6)$
$\frac{3}{2}\times 2\times 1\times 2192.4$
$=6577.2cal$
Since, process involve cooling$\therefore q_v=\Delta U=-6577.2cal$
Also in path AB, the internal energy in state A and state C is same. Thus during path AB, an increase in internal energy equivalent of change in internal energy during path BC should take place.Thus,$\Delta U$ for path AB=+6577.2cal
Now q for path $AB=\Delta U-w_{AB}=6577.2+4384.9=10962.1cal$
Cycle:$\Delta U=0:q=-w=-{[w}_{PathCA}+w_{PathAB}+w_{PathBC}]$
$=-[+1122.02+(-4384.9)+0]$
$\therefore q=-w=+3262.88cal$