Question

Consider a `P-V` diagram in which the path followed by one mole of perfect gas in a cyclinderical 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 temperatures `T_(1)//T_(2), if V_(2)= 2V_(1)`?
(c ) Given the internal energy for one mole of gas at temperature `T is (3//2) RT`, find the heat supplied to the gas when it is taken from state 1 to 2, with `V_(2)= 2V_(1)`.

Answer 1

Let `pV^(1//2)` = Constant = `K, p=(K)/(sqrt(V))`
(a) Work done for the process 1 to 2, ltBrgt `W=int_(V_(1))^(V_(2))pdV=Kint_(V_(1))^(V_(2))(dV)/(sqrt(V))=K[(sqrt(V))/(1//2)]_(V_(1))^(V_(2))=2K(sqrt(V_(2))-sqrt(V_(1)))`
`=2p_(1)V_(1)^(1//2)(sqrt(V_(2))-sqrt(V_(1)))=2p_(2)V_(2)^(1//2)(sqrtV_(2)-sqrt(V_(1)))`
(b) From ideal gas equation,
`pV=nRTrArrT=(pV)/(nR)=(psqrt(V)sqrt(V))/(nR)`
`rArr" "T=(Ksqrt(V))/(nR)" "("As",psqrt(V)=K)`
Hence, `" "T_(1)=(Ksqrt(V_(1)))/(nR)rArrT_(2)=(Ksqrt(V_(2)))/(nR)`
`rArr" "(T_(1))/(T_(2))=((Ksqrt(V_(1)))/(nR))/((Ksqrt(V_(2)))/(nR))=sqrt(V_(1)/(V_(2)))=sqrt(V_(1)/(2V_(1)))=(1)/(sqrt(2))" "(because V_(2)=2V_(1))`
(c ) Given, internal energy of the gas = `U=((3)/(2))RT`
`" "DeltaU=U_(2)-U_(1)=(3)/(2)R(T_(2)-T_(1))`
`" "=(3)/(2)RT_(1)(sqrt(V)-1)" "[becauseT_(2)=sqrt(2)T_(1)"from"(b)]` ltbgt `" "DeltaW=2p_(1)V_(1)^(1//2)(sqrt(V_(2))-sqrt(V_(1)))` ltBrgt `" "=2p_(1)V_(1)^(1//2)(sqrt(2)xxsqrt(V_(1))-sqrt(V_(1)))`
`" "=2p_(1)V_(1)(sqrt(2)-1)=2RT_(1)(sqrt(2)-1)`
`because " " DeltaQ=DeltaU+DeltaW`
`" "=(3)/(2)RT_(1)(sqrt(2)-1)+2RT_(1)(sqrt(2)-1)`
`" "=(sqrt(2)-1)RT_(1)(2+3//2)`
`" "=((7)/(2))RT_(1)(sqrt(2)-1)`
This is the amount of heat supplied.