In the Thermodynamic Process of Decorating

Handout solutions:

5. We need to convert to Pascals and m^3: 4.0 L = 4 x 10^-3m^3 and 2.0 L = 4 x 10^-3m^3.

Since the initial gauge pressure is 3 atm, the absolute pressure is 4 atm = 405,200 Pa.

For a constant pressure process, W = -P*delta_V = -(405,200 Pa)(-2 x 10^-3m^3) = 810.4 J

b) Since it's a constant pressure process, the final pressure is also 405,200 Pa.

c) Using ideal gas law, T2 = T1*(V2/V1) = 149 K

6.
a) Using ideal gas law, 5P0*3V0 = nRT0 so T0 = 15P0V0/nR

b) Since temperature is constant, the change of internal energy is zero.

c) Using ideal gas law again, P2V2 = nRT2

We know that T2 = T0 = 15P0V0/nR so P2V2 = 15P0V0.

V2 = 6V0

P2 = 15P0V0/6V0 = 5/2P0

7.
a) Ideal gas law - n = PV/nT = (7*101,300Pa)(10 x 10^-3m^3)/(8.315.J/Mol*K * 500 K) = 1.706 moles

b) The container volume is constant, so the work done is zero.

c) T2 = T1*P2/P1 = 214.3 K

d) THe graph should be a vertical line at V = 10 x 10^-3 m^3 from P = 7 atm to 3 atm (or the equivalent pressures in Pa.)

Book HW:

3,4,10abc, 53a,c

3.

4.

10
a) and c) PV diagram:

b) W = -2700 J, found since the process is isobaric, using W = -P*delta_V

53.
a) W = -2.2 x 10⁵ J

c)