Starting with volume and temperature of 9.2 x 10 m³ and 290 K, respectively, 0.183 [6] moles of an ideal 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 m³, isochoric heating to 420 K, isobaric expansion to 4.4 x 10 m³, and adiabatic cooling back to the original state.

Question

a. Define each of the four processes that the gas goes through.

[8]

b. Evaluate the missing state variables (P.V.T) at the end of each thermodynamic process.

[10]

c. Neatly sketch a PV diagram for the cycle indicating actual values.

Answer 1

a. The four processes that the gas goes through in the thermodynamic cycle are:

1. Isothermal compression: In this process, the gas is compressed while maintaining a constant temperature. The gas volume decreases, but the pressure increases.
2. Isochoric heating: In this process, the gas is heated while the volume remains constant. The pressure and temperature of the gas increase.
3. Isobaric expansion: In this process, the gas expands while the pressure remains constant. The volume of the gas increases, and the temperature decreases.
4. Adiabatic cooling: In this process, the gas undergoes a decrease in temperature without any heat entering or leaving the system. The volume decreases, and the pressure decreases as well.

b. To evaluate the missing state variables at the end of each process, we use the ideal gas law equation:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature

At the end of each process, we can calculate the missing state variables:

1. Isothermal compression:
Initial volume (V1) = 9.2 x 10 m³
Initial temperature (T1) = 290 K
Final volume (V2) = 2.3 x 10 m³

Using the ideal gas law equation:
P1V1/T1 = P2V2/T2

We can solve for P2:
P2 = (P1V1/T1) * (T2/V2)
P2 = (P1V1 * T2) / (T1 * V2)

2. Isochoric heating:
Pressure (P2) = Value obtained from the previous calculation
Initial temperature (T2) = Value obtained from the previous calculation
Final temperature (T3) = 420 K

Using the ideal gas law equation:
P2V2/T2 = P3V3/T3

We can solve for P3:
P3 = (P2V2/T2) * (T3/V3)
P3 = (P2V2 * T3) / (T2 * V3)

3. Isobaric expansion:
Pressure (P3) = Value obtained from the previous calculation
Final volume (V3) = 4.4 x 10 m³
Initial temperature (T3) = Value obtained from the previous calculation

Using the ideal gas law equation:
P3V3/T3 = P4V4/T4

We can solve for T4:
T4 = (P3V3/T3) * (T4/V4)
T4 = (P3V3 * T4) / (T3 * V4)

4. Adiabatic cooling:
Initial volume (V4) = 4.4 x 10 m³
Initial temperature (T4) = Value obtained from the previous calculation
Final temperature (T1) = 290 K

Using the ideal gas law equation:
P4V4/T4 = P1V1/T1

We can solve for P4:
P4 = (P1V1/T1) * (T4/V4)
P4 = (P1V1 * T4) / (T1 * V4)

c. To sketch the PV diagram for the cycle, we plot pressure (P) on the y-axis and volume (V) on the x-axis. The actual values obtained from the calculations in part b are then plotted on the diagram for each process.