PreQuiz 15.1 - PV Diagrams .:. Go Up

The Formulas: W = work done changing the volume in J

P = Pressure in Pascals (1 Pa = 1 N/m2)  Gauge pressure (PG) is the amount of pressure more than one atmosphere.  P = PG + 1 ATM
DV = change in volume in m3. P = the absolute pressure in Pa.  Don't use gauge pressure.  If you get pressure in psi, or any other unit, and or gauge, convert it to Pa absolute:
4.0 psi gauge = 4.0 psi + 14.7 psi = 18.7 psi absolute (because 1 atm in psi is 14.7)  and finally, 18.7/14.7 = P/1.013x105,  so P = 126,797.2789 Pa absolute.
Remember, 1 atm = 1.013x105 Pa = 101.3 kPa = 14.7 psi = 760 Torr.
V = Volume in cubic meters.  (m3)  There are 1000 liters in 1 m3.
n = number of mols.  n = mass/molar mass.
R =
8.31JK-1mol-1
T = Absolute temperature in KELVINS.  T = oC + 273.15

The Prequiz:

This is a Pressure v Volume graph for 0.28 mols of an ideal gas. The system starts at a pressure of 175. Pa and a volume of 1.60 m3 and goes through these four processes:

1. Isobaric (constant pressure) compression to 0.40 m3
2. Isochoric (constant volume) cooling to 50. Pa
3. Isobaric expansion to 1.60 m3
4. Isochoric heating to 175 Pa

Draw all four processes.  Use arrows for the processes, and label each process 1, 2, 3 or 4

1. Calculate the temperature (in K) at the end of each process.  Label these temperatures at the corners on the graph above.

Starting at the beginning of cycle 1:

 P = 175 Pa, V = 1.60 m3: PV = nRT, (175 Pa)(1.60 m3) = (0.28 mols)(8.31 J/mol K)(T) - the mols are given above the graph P = 175 Pa, V = 0.40 m3: PV = nRT, (175 Pa)(0.40 m3) = (0.28 mols)(8.31 J/mol K)(T) P = 50.0 Pa, V = 0.40 m3: PV = nRT, (50.0 Pa)(0.40 m3) = (0.28 mols)(8.31 J/mol K)(T) P = 50.0 Pa, V = 1.60 m3: PV = nRT, (50.0 Pa)(1.60 m3) = (0.28 mols)(8.31 J/mol K)(T)

2. Which processes have Zero net work done?  Why is no work done? (Answer in words)

2 and 4 – the volume does not change

3. Which processes have non-zero work?  What is the work done by each process that has non-zero work?

Processes 1 and 3 have a change in volume, so we can use to calculate the work:

Process 1: W = (175 Pa)(0.40 m3  - 1.60 m3) = -210. J (Negative because the change in volume was negative)
Process 3: W = (50.0 Pa)(1.60 m3 - 0.40 m3) = + 60.0 J

4. What is the net overall work done by this cycle?

Um -210 J + 60.0 J = -150. J