Question

1 litre of an ideal gas $(\gamma =1.5)$ at 300 K is suddenly cpmpressed to half its original volume.

(a) Fibd the ratio of the final pressure to the initial pressure.

(b) If the original pressure is 100 kPa, find the work done by the gas in the process.

(c) What is the change in internal energy ?

(d) What is the final temperature ?

(e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done durting the process.

(f) The gas is now expanded isothermally to achieve its original volume of 1 litre Calculate the work done by the gas

(g) Calculate the total work done in the cycle.

Answer 1

$\gamma =1.5,T=300K,V_1=1L,V_2=\frac{1}{2}L$

(a) The process is adiabatic because volume is suddenly change

$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

or $P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }=P_1(2{)}^{\gamma }$

or $\frac{P_2}{P_1}=2^{1.5}=2\sqrt{2}$

(b) $P_1=100KPa={10}^{5}Pa$

and $P_2=2\sqrt{2}\times {10}^{5}Pa$

Work done by adiabatic process

= $\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}$

= $\frac{{10}^5\times {10}^{-3}-2\sqrt{2}\times {10}^5\times \frac{1}{2}\times {10}^{-3}}{1.5-1}$

= - 82 J

(c) Internal energy,

dQ = 0

$\Rightarrow dU=-dW=-(-82J)=82J$

(d) $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

$T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$

= $300(2{)}^{0.5}$

= $300\times \sqrt{2}\times =300\times 1.414$

$T_2$ = 424 K

(e) The pressure is kept constant. The process is isobaric work done = nRdT,

Here, $n=\frac{PV}{RT}=\frac{{10}^5\times {10}^{-3}}{R\times 300}=\frac{1}{3R}$

So work done = $\frac{1}{3R}\times R(300-424)$

= -41.4 J.

(f) $\frac{V_1}{T_1}=\frac{V_2}{T_2}.....\left(1\right)$

$V_1=V_2\frac{T_1}{T_2}$

Work done in this process

= $nRTIn\frac{V_1}{V_2}$

= $\frac{1}{3R}\times R\times T\times In2$

= 100 $\times In2=100\times 1.039$

= 103

(g) Net Work done

= -82 -41.4 +103

= -20.4J.