a.
Process 1: Isothermal compression
Using the ideal gas law, PV = nRT:
P1V1 = nRT1
P2V2 = nRT1
Since the process is isothermal:
T1 = T2 = 290 K
V1 = 9.2 x 10^-4 m^3
V2 = 2.3 x 10^-4 m^3
n = 0.183 moles
R = 8.31 J/mol*K
Solving for P2:
P2 = (nRT1)/V2
P2 = (0.183 * 8.31 * 290)/2.3x10^-4
P2 = 508726 Pa
Process 2: Isochoric heating
Since the volume is constant, the pressure must increase to achieve a rise in temperature:
V2 = 2.3 x 10^-4 m^3
T2 = 420 K
n = 0.183 moles
R = 8.31 J/mol*K
Using the ideal gas law again:
P1V1 = nRT1
P2V2 = nRT2
Since the volume is constant:
V1 = V2
P1 = (nRT1)/V1
P2 = (nRT2)/V1
Since P1 = P2 in this process:
P2 = (nRT1)/V1
P2 = (0.183 * 8.31 * 290)/2.3x10^-4
P2 = 508726 Pa
Process 3: Isobaric expansion
The pressure is constant in this process, so we can use the ideal gas law to find the final volume:
P2 = 508726 Pa
T2 = 420 K
n = 0.183 moles
R = 8.31 J/mol*K
V1 = 2.3 x 10^-4 m^3
P1V1 = nRT1
P2V2 = nRT2
Since P1 = P2:
V2 = (nRT2)/P2
V2 = (0.183 * 8.31 * 420)/508726
V2 = 1.87 x 10^-4 m^3
Process 4: Adiabatic cooling
Since the process is adiabatic, we can use the adiabatic relation:
PV^γ = constant
Since this is a monoatomic gas, γ = 5/3:
P2V2^5/3 = P1V1^5/3
P1 = P2(V2/V1)^5/3
P1 = 508726(1.87x10^-4/9.2x10^-4)^5/3
P1 = 41343 Pa
b.
The PV diagram looks like this:
[Diagram not currently visible]
c.
Process 1: Isothermal compression
i. Work: W = nRT1ln(V1/V2) = -83.7 J
ii. Internal energy change: ΔU = 0 (because the temperature is constant)
iii. Entropy change: ΔS = nRln(V1/V2) = -0.27 J/K
Process 2: Isochoric heating
i. Work: W = 0 (because the volume is constant)
ii. Internal energy change: ΔU = (3/2)nR(T2 - T1) = 51.8 J
iii. Entropy change: ΔS = (3/2)nRln(T2/T1) = 0.64 J/K
Process 3: Isobaric expansion
i. Work: W = PΔV = 71.5 J
ii. Internal energy change: ΔU = (3/2)nR(T2 - T1) = 51.8 J
iii. Entropy change: ΔS = (3/2)nRln(T2/T1) = 0.64 J/K
Process 4: Adiabatic cooling
i. Work: W = nCv(T1 - T2) = -36.5 J
ii. Internal energy change: ΔU = (3/2)nR(T1 - T2) = -31.4 J
iii. Entropy change: ΔS = 0 (because the process is adiabatic)
Starting with volume and temperature of 9.2 x 10-4 m3
and 290 K, respectively, 0.183
moles of an ideal monoatomic gas contained in a cylinder with a moveable frictionless
piston undergoes the following order of processes in a thermodynamic cycle: isothermal
compression to a volume of 2.3 x 10-4 m3
, isochoric heating to 420 K, isobaric expansion
to 4.4 x 10-4 m3
, and adiabatic cooling back to the original state.
a. Evaluate the missing P,V,T at the end of each thermodynamic process
b. Sketch the corresponding PV diagram indicating the applicable P,V values
c. For each process, find the corresponding:
i. Work
ii. Internal energy change
iii. Entropy change
1 answer