problem1. A sample consists of 1.0 mole of an ideal monatomic gas (Cv=3/2R, U=3/
ID: 887340 • Letter: P
Question
problem1. A sample consists of 1.0 mole of an ideal monatomic gas (Cv=3/2R, U=3/2RT, Cp=5/2R, R = 8.315JK^-1M^-1, 1 Lit atm = 101 J). It is taken through the cycle below.
Between points 2 and 1 the temperature is constant at T=300K. At point A the volume is 24.6 Liters and the pressure is 1.00 atm.
Between point 1 and 4 the pressure is constant at 0.5 atm.
Between point 4 and 3 the temperature is constant at T=600K.
Between points 3 and 2 the pressure is constant at 1atm.
problem 2. Calculate q, w, delta U, delta H for each step. Leave you answer in terms of R, the gas constant, or in joules. using the sigh (+,-,0)
step 2-1
step 1-4
step 4-3
step 3-2
what is the net heat around the whole cycle?
what is the net work around the whole cycle?
what is the overall internal enrgy change delta U around the whole cycle ?
quantity sign Numerical ans. quantity sign Numerical ans. q w delta U delta HExplanation / Answer
From 1 to 4 pressure constant = 0.5 atm
dU = nCvdT
= 1 x 3/2 x 8.315 x (600-300)
= 3.74 kJ
w = nRdT
= 1 x 8.315 x 300
= 2.50 kJ
q = nCpdT
= 1 x 5/2 x 8.315 x 300
= 6.24 kJ
H = q = 6.24 kJ
From 3 to 2, pressure constant = 1 atm
dU = nCvdT
= 1 x 3/2 x 8.315 x (-300)
= -3.74 kJ
w = nRdT
= 1 x 8.315 x -300
= -2.50 kJ
q = nCpdT
= 1 x 5/2 x 8.315 x -300
= -6.24 kJ
H = q = -6.24 kJ
From 2 to 1
P1V1 = P2V2
with, P1 = 1.0 atm ; V1 = 24.6 L
P2 = 0.5 atm ; V2 = unknown
V2 = 1 x 24.6/0.5
= 49.2 L
w = -nRTln(Vf/Vi)
= - 1 x 8.315 x 300 x ln (2)
= -1.73 kJ
When dT = const. ; dU = 0
So, q = -w = 1.73 kJ
H = -w = 1.73 kJ
From 4 to 3,
w = -nRTln(Vf/Vi)
= - 1 x 8.315 x 600 x ln (0.5)
= 3.46 kJ
When dT = const. ; dU = 0
So, q = -w = -1.73 kJ
H = -w = -1.73 kJ
So, the net heat around the cycle = 0 kJ
Net work done around the cycle = -1.73 + 3.46 = 1.73 kJ
Overall internal energy change dU = 0 kJ
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