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

Answer:

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

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 ; $PV^{1.27} = c$

$\dfrac{T_2}{T_1}= ( \dfrac{V_1}{V_2})^{n-1} = (\dfrac{P_2}{P_1})^{\frac{n-1}{n}$

$T_2 = T_1 [\dfrac{P_2}{P_1}]^{\frac{n-1}{n}$

$T_2 = 303 [\dfrac{100}{200}]^{\frac{1.27-1}{1.27}$

$T_2 = 261.48 \ K$

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

$W = \dfrac{P_1V_1-P_2V_2}{n-1} = \dfrac{mR(T_1-T_2)}{n-1}$

$W = \dfrac{1*188.9(303-261.48)}{1.27-1}$

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