Two moles of helium gas`(lambda=5//3)` are initially at temperature `27^@C` and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value.
(i) Sketch the process on a p-V diagram.
(ii) What are the final volume and pressure of the gas?
(iii) What is the work done by the gas ?

Explanation

a. From ideal gas equation

`PV=nRT`
Initial pressure
`PV=nRT=(2xx8.3xx300)/(20xx10^(-3))=2.49xx10^(5)N//m^(2)`
When volume o gas is doubled at contant pressure, its temperature is also doubled. This process is shown on P-V curve by line AB. The gas then cools to temperatrue T adiabatically. This is shown by curve BC. The whole process is represented by curve ABC.
b. At point B, pressure `P_(B)^(.)=P_(A)=2.49xx10^(5)N//m^(2)`
Volume `V_(B)=2V_(A)=40xx10^(-3)m^(3)`, Temperature `T_(B)=600K`
now from adiabatic equation`TV^(r-1)=` constant
We have `T_(B)V_(B^((gamma-1))=T_(C)V_(C)^((gamma-1))`
`:>((V_(c))/(V_(B)))^(gamma-1)=(T_(B))/(T_(C))=600/300=2`
`:.(V_(C))/(V_(B))=2^(1//(gamma-1))=2^(3//2)`
Final volume
`:.V_(C)=2sqrt(2)V_(B)=2xx1.414xx40 l =113l`
and final pressure
`P_(C)=(nRT_(C))/(V_(C))=(2xx8.3xx300)/(113.13xx10^(-3))=0.44xx10^(5)N//m^(2)`
c. The work done by gas is isobaric process AB
`=2.49xx10^(5)xx(40-20)xx10^(-3)=4980J`
The work done by gas during adiabatic process BC
`W_(2)=(nR)/(1-gamma)[T_(2) -T_(1)]=(2xx8.3)/(1-(5//3))[300-6300]=7470J`
`:.` Net work done by gas `W=W_(1)+W_(2)=4980+7470=12450J`