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)`.

Explanation

(a) As shown in (figure)
Here, `PV^(1//2)= constant = A`, say
As `W= int_(V_(1))^(V_(2)) pdV :. W=int_(V_(1))^(V_(2)) A/(V^(1//2)) dV= A[(V^(1//2))/(1//2)]_(V_(1))^(V_(2))= 2A [sqrt(V_(2))-sqrt(V_(2))]= 2P_(1)V_(1)^(1//2) [sqrt(V_(2))-sqrt(V_(1))]`..(i)
(b) From standard gas equation, `(PV)/T= nR :. T(PV)/(nR)= (Asqrt(V))/(nR) prop sqrt(V)`
`(T_(2))/(T_(1))= sqrt(V_(2)/V_(1))= sqrt(2)`
(c ) As `U=3/2RT, DeltaU= U_(2)-U_(1)=3/2R(T_(2)-T_(1))`
Using (ii) `DeltaU= 3/2RT_(1)(sqrt(2)-1)`
When `V_(2)= 2V_(1)`, from, (i),
`DeltaW=2P_(1)V_(1)^(1//2)(sqrt(V_(2))-sqrt(V_(1)))= 2P_(1)V_(1)^(1//2) sqrt(V_(1))(sqrt(2)-1)= 2P_(1)V_(1)(sqrt(2)-1)= RT_(1)(sqrt(2)-1)`
As `DeltaQ= DeltaU+DeltaW, :. DeltaQ= 3/2 RT_(1)(sqrt(2)-1)+ 2RT_(1)(sqrt(2)-1)= 7/2RT_(1)(sqrt(2)-1)`