220 kg of CO_{2} gas at 27^{o}C and 1 atm is compressed adiabatically to 1/5th of its volume. It is then cooled to its original temperature at constant volume. Find Q, ΔU and W for each step and for the entire process.
Formula:
For an ideal gas, PV = nRT
For an adiabatic process, PV^{γ} = constant,
V_{2}/V_{1} = (P_{1}/P_{2})^{(1/γ) }
T_{2}/T_{1} = (V_{1}/V_{2})^{(γ1) }
Adiabatic work of compression, W = (P_{2}V_{2}  P_{1}V_{1}) / (γ  1) → 1
Q + W = ΔU
ΔU = mC_{V}(T_{2}  T_{1}) → 2
Calculations:
Here there are two steps taking place.
n = 220 / Molecular weight of CO_{2} = 220/44 = 5 kmol
V_{1} = 5 x 8314 x (273 + 27) / (1.01325 x 10^{5}) = 123.08 m^{3}.
V_{2} = (1/5) x 123.08 = 24.616 m^{3}.
γ = 1.3 (for CO_{2}: data)
T_{2} = 300 x (5)^{0.3} = 486.2 K
P_{1}/P_{2} = (V_{2}/V_{1})^{γ}
P_{2} = 1 / (1/5)^{γ}
i.e., P_{2} = 8.103 atm.
Adiabatic compression work W can be calculated from equn.1 or equn.2 (since ΔU = Q + W; and Q = 0 for an adiabatic process).
W = (8.103 x 1.01325 x 10^{5} x 24.616  1.01325 x 10^{5} x 123.08) / 0.3
= 25.7985 MJ
Since the temperature of the system is returned to its original state, the internal energy change of the total process is zero. Therefore, ΔU for the heat removal step is = 25.7985 MJ.
For the constant volume cooling step, W = 0, and ΔU = 25.7985 MJ. Therefore, Q for this step = 25.7985 MJ.
Summary:
Q 
W 
ΔU 

Step 1 
0 
25.7985 MJ 
25.7985 MJ 
Step 2 
25.7985 MJ 
0 
25.7985 MJ 
Over all 
25.7985 MJ (heat removed from the system) 
25.7985 MJ (work added to the system) 
0 
Last Modified on: 04Feb2022
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