Consider the expansion of 10.0 L of gas originally at 10.0 atm pressure to a fin
ID: 573609 • Letter: C
Question
Consider the expansion of 10.0 L of gas originally at 10.0 atm pressure to a final pressure of 1.0 atm. The T remains constant at 250°C, Defining the gas as the system, calculate the work done when a.the P on the gas is suddenly decreased to 1.0 atm. Answer:-90 L atm b.the P. on the gas is suddenly decreased to 5.0 atm, allowed to come to equilibrium, and then the P. is again suddenly decreased to 1.0 atm. Answer:-130 L atm c.the P. on the gas is suddenly decreased to 6.67 atm, allowed to come to equilibrium, and then the P. is again suddenly decreased to 3.33 atm, allowed to come to equilibrium, and, finally, the P. is suddenly decreased to 1.0 atm. Answer:-153 L atmExplanation / Answer
a) Since the pressure is suddenly dropped to 1 atm, the process is irreversible and the work done in this case will be
W = - Pext (V2 - V1) where Pext = 1.0 atm V1 = 10.0 L. We have to calculate volume V2 first
P1V1 = P2V2
(10.0 atm)x(10.0 L) = (1.0 atm)xV2;
V2 = 100 L
W = -(1 atm)x(100 L - 10 L)
= -90 L-atm
b)In this case the process takes place in 2 steps both being irreversible:
Step 1: When pressure is suddenly dropped to 5.0 atm
W1 = - Pext (V2 - V1) where Pext = 5.0 atm V1 = 10.0 L. We have to calculate volume V2 first
P1V1 = P2V2
(10.0 atm)x(10.0 L) = (5.0 atm)xV2;
V2 = 20 L
W1 = -(5 atm)x(20 L - 10 L)
= -50 L-atm
Step 2: When pressure is suddenly dropped to 1.0 atm from 5.0 atm
W2 = - Pext (V3 - V2) where Pext = 1.0 atm V2 = 20.0 L. We have to calculate volume V3 first
P2V2 = P3V3
(5.0 atm)x(20.0 L) = (1.0 atm)xV3;
V3 = 100 L
W2 = -(1 atm)x(100 L - 20 L)
= -80 L-atm
Therefore, total work done W = W1 + W2 = -50 + (-80) L-atm = -130 L-atm.
c) In this case the process takes place in 3 steps all being irreversible:
Step 1: When pressure is suddenly dropped to 6.67 atm
W1 = - Pext (V2 - V1) where Pext = 6.67 atm V1 = 10.0 L. We have to calculate volume V2 first
P1V1 = P2V2
(10.0 atm)x(10.0 L) = (6.67 atm)xV2;
V2 = 15 L
W1 = -(6.67 atm)x(15 L - 10 L)
= -33.35 L-atm
Step 2: When pressure is suddenly dropped to 3.33 atm from 6.67 atm
W2 = - Pext (V3 - V2) where Pext = 3.33 atm V2 = 15.0 L. We have to calculate volume V3 first
P2V2 = P3V3
(6.67 atm)x(15.0 L) = (3.33 atm)xV3;
V3 = 30.05 L
W2 = -(3.33 atm)x(30.05 L - 15 L)
= -50.12 L-atm
Step 3: When pressure is suddenly dropped to 1 atm from 3.33 atm
W3 = - Pext (V4 - V3) where Pext = 1 atm V3 = 30.05 L. We have to calculate volume V4 first
P3V3 = P4V4
(3.33 atm)x(30.05 L) = (1 atm)xV4;
V4 = 100.07 L
W3 = -(1 atm)x(100.07 L - 30.05 L)
= -70.02 L-atm
Therefore, total work done W = W1 + W2 + W3 = -33.35 + (-50.12) + (-70.02) L-atm = -153.5 L-atm.