A gas contained in a piston-cylinder assembly undergoes two processes, A and B,

Question

A gas contained in a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where p1= 10 bar, V1= 0.1 m^3, U1= 400 kJ and P2= 1 bar, V2= 1 m^3, U2= 200 kJ: Process A: Constant-volume process from state 1 to a pressure of 10 bar, followed by a constant-pressure process to state 2. Process B: process from 1 to 2 during which the pressure-volume relation is PV= Constant. Kinetic and Potential energy effects can be ignored.

For each of these processes A and B: (a.) evaluate the work, in kJ, (b.) evaluate the heat transfer, in kJ.

Answers are: For process A ---> W = 230.26 kJ, Q = 30.26 kJ

For process B ----> W = 135 kJ, Q = -65 kJ

Explanation / Answer

Work is given by the integral
W = - p dV along the path
If you choose a different path. i.e. a different relation between p and V the work change. Q on this path changes too, giving overall the same change of internal energy:
U = Q + W

The change of internal energy is the same in both cases
U_A = U_B = U - U = 200kJ - 400kJ = -200kJ

For process A
pV = constant = pV
hence:
p = pV / V
and
W_A = - pV 1/V dV from V to V
= - pV ln(V/V)
= - 10^6Pa 0.1m³ ln(1m³ / 0.1m³)
(since Pam³ = J)
= - 10^5J ln(10)
-2.3025×10^5J = -230.25kJ
=>
Q_A = U_A - W
= -200kJ - (-230.25kJ) = 30.25kJ

In first step of process no work is done because the volume does not change. So the work is entirely determined by the second step.
I t follows a linear path in the p,V diagram from
p'=10 bar, V=1m³ to p=1bar, V=1m³

The pressure on the path is given by
p = p' + [(p - p')/(V - V)](V - V)
So the work on this path is
W_B = - p' + [(p - p')/(V - V)](V - V) dV from V to V
= - { p'(V - V) + [(p - p')/(V - V)] ((1/2)(V² - V²) - V(V - V)) }
= - { p'(V - V) + [(p - p')/(V - V)] (1/2) (V² - 2VV + V²) }
= - { p'(V - V) + [(p - p')/(V - V)] (1/2) (V - V)² }
= - { p'(V - V) + (1/2)(p - p')(V - V) }
= - (1/2) (p' + p) (V - V)
= - (1/2) (10 ×10^5Pa + 1.0×10^5Pa) (1m³ - 0.1m³)
= - 4.95 ×10^5J = -495kJ
=>
Q_B = U_B - W
= -200kJ - (-495kJ) = 295kJ

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