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