Question

Two mole of helium gas $(\gamma =5/3)$ are initially at a temperature of ${27}^{o}C$ and occupy a volume of 20 litre. the gas is first expanded at constant pressure until the volume is doubled. it then undergoes adiabatic change until the temperature return to its initial value.
(a) Sketch the process on P-V disgram.
(b) What are the final pressure and final volume of gas?
(c) What is the work done by the gas?

Answer 1

The PV diagram for the given changes is as:
at A: $T_1=300K,n=2,V=20litre,P=P_1$
at B: $T=T_2,n=2,V=40litre,P=P_1$
at C: $T=300K.n=2,V=V_{1}litre,P=P_2$
For A $P_1=\frac{nRT}{V}=\frac{2\times 0.082\times 300}{20}=2.46atm$
For isobaric process AB:
Also $\frac{V_1}{T_1}=\frac{V_2}{T_2}\left(fromCharle{s}^{\prime }lawatconstantP\right)$
$\frac{20}{300}=\frac{40}{T_2}$
$\therefore T_2=600K$
For adiabatic process BC:
${TV}^{y-1}=constant$
$600\times (40{)}^{y-1}=300\times (V_1{)}^{y-1}$
or $(\frac{V_1}{40}{)}^{\left(\frac{5}{3}1\right)}=2$
or $(\frac{V_1}{40}{)}^{2/3}=2$
$\therefore V_1=113.13litre$
The work done $W_T=W_{AB}+W_{BC}$
$W_{AB}:\therefore w=nRT.dT$
$w=2\times 2\times (600-300)=1200cal$
$W_{BC}:foradiabaticprocess;$
$\therefore w=-{nC}_v.\Delta T$
$=-2\times \frac{3}{2}\times R\times (600-300)$
$=1800cal$
$\therefore W_r=1200+1800=3000cal$