Question

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

1. Find the work done when the gas is taken from state 1 to state 2.
2. What is the ratio of temperature T₁/T₂, if V₂ = 2V₁?
3. 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 V₂ = 2V₁.

Answer 1

Let `PV^(1/2) = K` = constant, `P = K/sqrt(V)`

Hence we can write, `P_1V_1^(1/2) = P_2V_2^(1/2) = K`

a. Work done for the process 1 to 2,

`W_(1 - > 2) int_(V_1)^(V_2) PdV`

= `int_(V_1)^(V_2) K/sqrt(V) dV`

= `K[sqrt(V)/(1/2)]_(V_1)^(V_2)`

= `2K(sqrt(V_2) - sqrt(V_1))`

= `2P_1V_1^(1/2) (sqrt(V_2) - sqrt(V_1))`

= `2P_2V_2^(1/2) (sqrt(V_2) - V_1)`

b. From the ideal gas equation,

`PV = nRT`

⇒ `T = (PV)/(nR)`

= `(Psqrt(V) sqrt(V))/(nR)`

⇒ `T = (Ksqrt(V))/(nR)`  .....As, `P sqrt(V) = K`)

i.e., `T ∝ sqrt(V)`

Thus, `T_2/T_1 = sqrt(V^2/V_1)`

= `sqrt((2V_1)/V_1)`

= `sqrt(2)`  .....(As V₂ = 2V₁)

c. Given, the internal energy of the gas, `U = (3/2) RT`

`ΔU = U_2 - U_1`

= `3/2 R(T_2 - T_1)`

= `3/2 R(sqrt(2)T_1 - T_1)`

= `3/2 RT_1(sqrt(2) - 1)`  .....[∵ T₂ = `sqrt(2)` T₁ from (b)]

`ΔW = W_(1 -> 2)`

= `2P_1V_1^(1/2) (V_2^(1/2) - V_1^(1/2))`

= `2P_1V_1 (V_1^(1/2)/V_1^(1/2) - 1)`

= `2RT_1(sqrt(2) - 1)`  ......`("As" P_1V_1 = RT_1 and V_2^(1/2)/V_1^(1/2) = sqrt(2))`

`ΔQ = ΔU + ΔW = 3/2 RT_1 (sqrt(2) - 1) + 2RT_1 (sqrt(2) - 1)`

= `7/2 RT_1 (sqrt(2) - 1)`

This is the amount of heat supplied.