Two moles of helium gas (`gamma = 5 / 3`) are initially at `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.
(a) Sketch the process in a `p_V` diagram.
(b) What is the final volume and pressure of the gas ?
(c) What is the work done by the gas?

Explanation

`n=2`,`gamma=(5)/(3)`,`T_1=T_0=27^@C=27+273=300K`
`V_1=V_0=20`litre`=20xx10^-3m^3`
`1rarr2:(V_0)/(T_0)=(2V_0)/(T_2)impliesT_2=2T_0`
`2rarr3:T_3V_2^(gamma-1)=T_3V_3^(gamma-1)`
`2T_0(2V_0)^(gamma-1)=T_0V_3^(gamma-1)`
`V_3=2^((1)/(gamma-1)).2V_0=2^((1)/((5//3-1))).2V_0=2^((3//2)).2V_0`
`=2^((5//2))V_0=4sqrt2V_B=4sqrt2xx20=80sqrt2` litre
`P_2V_2^(gamma)=P_3V_3^(gamma)impliesP_0(2V_0)^(gamma)=P_3(2^((5//3))V_0)^(gamma)`
`P_3=P_0((2V_0)/(2^((5//2))V_0))^(gamma)=P_0(2^((-3)/(2)))^((5)/(3))=2^((-5)/(2))P_0`
`=(P_0)/(4sqrt2)`
`P_0V_nRT_0`
`P_0xx20xx10^_3=2xx8.3xx300`
`P_0=24.9xx10^4(N)/(m^2)`
`P_3=(P_0)/(4sqrt2)=(24.9xx10^4)/(4sqrt2)=4.4xx10^4(N)/(m^2)`
(a)
(b) Final volume `=80sqrt2`litre
Final pressure`=4.4xx10^4(N)/(m^2)`
(c) `DeltaW_(1rarr2)=PDeltaV=P_0(2V_0-V_0)=P_0V_0=2RT_0`
`DeltaW_(2rarr3)=(nR(T_2-T_1))/(gamma-1)(2R(2T_0-T_0))/((5)/(3)-1)=(2RT_0)/((2)/(3))=3RT_0`
`DeltaW=DeltaW_(1rarr2)+DeltaW_(2rarr3)=5RT_0=5xx8.3xx300`
`=12450J`