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

For a perfect gas
`PV = nRT`
Given that `V = 20 L = 20 xx 10^(-3) m^(-3)`
`T = 27^(@)C = 3000 K` and number of molecules n = 2
Thus, initial pressure is given as
`P = ((nRT)/(V)) = ((2 xx 8.3 xx 300))/((20 xx 10^(-3))) = 2.5 xx 10^(5) N//m^(2)`
Figure 2.77 shows the inductior diagram of the complete process.
b. At point B,
Pressure P' ` = P = 2.5 xx 1-^(5) N//m^(2)`, and
`V' = 2V = 40 xx 10^(-3) m^(3)`
As pressure is constant in the process AB, making its volume doubled, its temperature will also be doubled.
Thus gas now undergoes adiabatic expansions to cool down at `T"b = T = 300 K`
We know for an adiabatic process `TV^(gamma - 1)` = constant
`T' (V')^(gamma - 1) = T" (V")^(gamma -1)`
`((V'')/(V')) = ((T')/(T''))^(1//(gamma - 1)) = ((600)/(300))^((1)/(2.5 -1)) = (2)^(3//2) = 2 sqrt2`
Thus, final volume is
`V" = (2 sqrt2) V'`
`2 xx 1/414 xx 40 xx 10^(-3) = 113.14 xx 10^(-3) m^(3)`
Similarly, final pressure is given by process equation as
`P' V'^(gamma) = P" V"^(gamma)`
or `P'' = P' ((V')/(V"))^(gamma)`
`= 2.5 xx 10^(5) xx ((40 xx 10^(-3))/(113.14 xx 10^(-3)))^(5//3)`
`4.42 xx 10^(4) Pa`
c. Work done under isobaric process AB is
or `W_(1) = P Delta V`
or `W_(1) = 2.5 xx 10^(5) xx (40 - 20) xx 10^(-3)`
`= 4980 J`
Work done during adiabatic process BC is given as
or `W_(2) ((nR)/(gamma - )) [T_(1) - T_(2)]`
or `= ((2 xx 8.3))/([1 - (5//3)]) [300 - 600] = 7470 J`
total work done `= W_(1) + W_(2) = 4980 + 7470 = 12450 J`