Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in figure.

(a) Find the work done when the gas is taken from state 1 to state 2.
(b) What is the ratio of temperature `T_1/T_2` if `V_2=2V_1` ?
(c) Given the internal energy for one mole of gas at temperature T is `3/2RT`, find the heat 2 supplied to the gas when it is taken from state 1 to 2, with `V_2 = 2V_1`

Explanation

From graph,
`PV^(1/2)`=constant K
`therefore P=K/sqrtV`
(a)Work done for the process 1 to 2,
`DeltaW=int_(V_1)^(V_2)P dV = int_(V_1)^(V_2)K/sqrtV xx dV`
`DeltaW=K[V^(1/2)/(1/2)]_(V_1)^(V_2) =2K[V_2^(1/2)-V_1^(1/2)]`…(1)
but `P_1V_1^(1/2)=K` and `P_2V_2^(1/2)=K`
`therefore W=2P_1V_1^(1/2)[V_2^(1/2)-V_1^(1/2)]` ....(2)
(b) From ideal gas equation of state ,
PV=nRT
`therfore T=(PV)/(nR)`
`therefore T=(PsqrtV.sqrtV)/(nR)`
`therefore T=(KsqrtV)/(nR)[ because PsqrtV=K]`
`therefore T_1=(KsqrtV_1)/(nR)` and `T_2=(KsqrtV_2)/(nR)`
`therefore T_1/T_2=sqrt(V_1/V_2)=sqrt((V_1)/(2V_1))[ because V_2=2V_1]`....(3)
`therefore T_1/T_2=1/sqrt2 therefore T_2=sqrt2T_1`...(4)
(c) Internal energy of gas ,
`U=3/2 RT`
`therefore DeltaU=U_2-U_1=3/2R(T_2-T_1)`
`DeltaU=3/2R(sqrt2T_1-T_1)=3/2RT_1(sqrt2-1)`....(5)
`rArr DeltaW=2P_1V_1^(1/2)[sqrtV_2- sqrtV_1]` (From equation 2)
`=2P_1V_1^(1/2)[sqrt(2V_1)-sqrtV_1]`
`=2P_1V_1[sqrt2-1]`
`=2RT_1[sqrt2-1][ because P_1V_1=RT_1]` ....(6)
From first rule of thermodynamics ,
`DeltaQ=DeltaU+DeltaW`
`=3/2RT_1(sqrt2-1)+2RT_1(sqrt2-1)`
`therefore DeltaQ=(7RT_1)/2 (sqrt2-1)`