Question

In a closed system one kilogram of carbon dioxide (CO_2) is expanded reversibly from 30 degree C and 200 kPa to 100 kPa pressure. If the expansion is polytropic with n = 1.27, determine the total work, the change in total internal energy, and the total heat transferred in [kJ], Note that for CO_2, R = 188.9 J/kg.K and c_v = 655 J/kg.K. W = -29.05 kJ, DeltaU = -27.19 kJ, Q = 1.860 kJ

Answer 1

the total work W = 29.05 kJ

the change in total internal energy is
`\mathbf{\Delta U = - 27.19 \ kJ}`

the total heat transferred in [kJ] is Q = 1.860 kJ

Step-by-step explanation:

Given that

mass of carbon dioxide in the closed system = 1 kg

Temperature
`T_1= 30 ^0 C` = (273+30 ) K = 303 K

Pressure
`P_1 = \ 200 \ kPa`

Pressure
`P_2 = 100 \ kPa`

polytropic expansion n = 1.27

Note that we are also given the following data set:

R = 188.9 J/kg.K

c_v = 655 J/kg.K

So; for a polytropic process ;
`PV^(1.27) = c`

`(T_2)/(T_1)= ( (V_1)/(V_2))^(n-1) = ((P_2)/(P_1))^{(n-1)/(n)`

`T_2 = T_1 [(P_2)/(P_1)]^{(n-1)/(n)`

`T_2 = 303 [(100)/(200)]^{(1.27-1)/(1.27)`

`T_2 = 261.48 \ K`

Since the system does not follow the first order of thermodynamics; To calculate the total work by using the expression:

`W = (P_1V_1-P_2V_2)/(n-1) = (mR(T_1-T_2))/(n-1)`

`W = (1*188.9(303-261.48))/(1.27-1)`

W = 29048.62222 J

W = 29.05 kJ

Thus, the total work W = 29.05 kJ

The change in internal energy can be expressed by the formula:

`\Delta U = mc_v (T_2-T_1)`

`\Delta U = 1*655(261.48-303)`

`\Delta U = -27195.6 \ J`

`\mathbf{\Delta U = - 27.19 \ kJ}`

Hence; the change in total internal energy is
`\mathbf{\Delta U = - 27.19 \ kJ}`

Finally; to determine the total heat transferred in [kJ]; we go by the expression for the first order of thermodynamics which say:

Total Heat Q = ΔU + W

Q = (-27.19 + 29.05)kJ

Q = 1.860 kJ

Hence; the total heat transferred in [kJ] is Q = 1.860 kJ