: A gas contained in a piston–cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where p1
=1bar, V1 = 1m3 , U1 =400kJ and p2 = 10 bar, V2 = 0.1 m3 , U2 = 450 kJ: Process A: Constant-volume process from state 1 to a pressure of 10 bar, followed by a constantpressure process to state 2. Process B: Process from 1 to 2 during which the pressure-volume relation is pV = constant. Kinetic and potential effects can be ignored. For each of the processes A and B, (a) sketch the process on p–V coordinates, (b) evaluate the work, in kJ, and (c) evaluate the heat transfer, in kJ.
1 answer:
Answer:
b) W = 9.09 10² kJ , W = -2.3 nRT
c) Q_total = -409 kJ , Q = 232 kJ
Explanation:
a) we have an outline of the two processes in the attached
Process A
Isocoric - Isobaric
B process
Isothermal
b) This is a thermodynamic process where we can find work with the expression
W = ∫P dV
A Process
In the constant volume part ΔV = 0, so the work is zero
W₁ = 0
In the constant pressure part
W = P DV
Let's reduce the pressure to SI units
P = 10 Bar (1.01 105 Pa / 1 Bar) = 10.1 10⁵ Pa
W₂ = 10.1 10⁵ (1 -0.1)
W₂ = 9.09 10⁵ J= 9.09 10² kJ
The total work is
W = W₁ + W₂
W = 9.09 10² kJ
B process
PV = d
Where d is a constant, we assume an ideal gas
PV = nRT
W = nRT ∫ dV / V
W = nRT ln V₂ / V₁
W = nrT ln 0.1 / 1
W = -2.3 nRT
c) Let's use the first law of thermodynamics
DU = Q-W
Process A
The first part of the process is at constant volume, so the work is zero
ΔU₁ = Q₁
Q₁ = 450 -400
Q₁ = 50 kJ
The second part
ΔU₂ = Q₂ + W₂
Q₂ = ΔU₂ - W₂
Q₂ = 450 - 909
Q₂ = -459 kJ
The total heat in the process is
Q_total = Q₁ + Q₂
Q_total = 50 - 459
Q_total = -409 kJ
The negative sign indicates that this heat is released
B process
ΔU = Q + W
Q = ΔU - W
Q = (450-400) - (-2.3 d)
Q = 50 + 2.3 d
let's use the initial values to calculate the constant
P V = d
1.01 10⁵ 1 = d
Q = 50 + 2.3 1.01 105
Q = 2.3 10⁵ J
Q = 2.323 10² kJ
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