*** THERMODYNAMICS *** 3. A closed piston contains 1.0 kg of ammonia at 20 C and
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Question
*** THERMODYNAMICS ***
3. A closed piston contains 1.0 kg of ammonia at 20 C and 100 kPa. The system is brought into contact with a heat reservoir at 50 C. The ammonia undergoes a polytropic process, with n = 1.084, until thermal equilibrium is reached.
(a) Find the final pressure, in kPa, assuming ideal gas behavior.
(b) When the experiment is carried out, the final pressure is actually 300 kPa. If your answer to part (a) was different, suggest why.
(c) Based on the actual final pressure of 300 kPa, calculate Qtotal and Wtotal, in units of kJ.
(d) Does this process satisfy the second law? [Hint: find the total change in entropy for the “universe.”]
Explanation / Answer
Using an ideal gas law, we have
P1 V1 = n R T1
V1 = [(1.084 mol) (8.314 J/mol.K) (293.15 K)] / (100 x 103 Pa)
V1 = (2641.9 J) / (100 x 103 Pa)
V1 = 0.0264 m3
From a charle's law, we have
V1 / T1 = V2 / T2
V2 = (0.0264 m3) [(50 0C) / (20 0C)]
V2 = 0.066 m3
(a) Assuming an ideal gas behavior, the final pressure will be given as :
using a gay-lussac law, we have
P1 / T1 = P2 / T2
where, P1 = initial pressure = 100 kPa
P2 = final pressure = ?
T1 = initial temperature = 20 0C
T2 = final temperature = 50 0C
Then, we get
P2 = (100 kPa) [(50 0C) / (20 0C)]
P2 = 250 kPa
(b) When the experiment is carried out, the final pressure is actually 300 kPa.
If our answer to part (a) was different, then it means that basically depends on temperature.
(c) Based on the actual final pressure of 300 kPa, then total work done will be given by -
we know that, Wtotal = Pfinal (V2 - V1)
Wtotal = (300 kPa) [(0.066 m3) - (0.0264 m3)]
Wtotal = (300 kPa) (0.0396 m3)
Wtotal = 11.8 kJ
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