**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/m

^{2}) Gauge pressure (P_{G}) is the amount of pressure more than one atmosphere. P = P_{G}+ 1 ATM

DV = change in volume in m^{3}.

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/1.013x10^{5}, so P = 126,797.2789 Pa absolute.

Remember, 1 atm = 1.013x10^{5}Pa = 101.3 kPa = 14.7 psi = 760 Torr.

V = Volume in cubic meters. (m^{3}) There are 1000 liters in 1 m^{3}.

n = number of mols. n = mass/molar mass.

R =8.31JKT = Absolute temperature in KELVINS. T =^{-1}mol^{-1 }^{o}C + 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 m^{3} and goes through these
four processes:

- Isobaric (constant
pressure) compression to 0.40 m
^{3} - Isochoric (constant volume) cooling to 50. Pa
- Isobaric expansion to 1.60
m
^{3} - 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 m ^{3}:

PV = nRT, (175 Pa)(1.60 m ^{3}) = (0.28 mols)(8.31 J/mol K)(T) - the mols are given above the graph

P = 175 Pa, V = 0.40 m ^{3}:

PV = nRT, (175 Pa)(0.40 m ^{3}) = (0.28 mols)(8.31 J/mol K)(T)

P = 50.0 Pa, V = 0.40 m ^{3}:

PV = nRT, (50.0 Pa)(0.40 m ^{3}) = (0.28 mols)(8.31 J/mol K)(T)

P = 50.0 Pa, V = 1.60 m ^{3}:

PV = nRT, (50.0 Pa)(1.60 m ^{3}) = (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 m

^{3 }- 1.60 m^{3}) = -210. J (Negative because the change in volume was negative)

Process 3: W = (50.0 Pa)(1.60 m^{3}- 0.40 m^{3}) = + 60.0 J

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

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