Figure 1 below are the pressure-volume and temperature-entropy diagrams for a four

Question

stroke Otto cycle which uses 1.67 moles of air as the working fluid (the gas that undergoes
the processes shown). The air has a molar specific heat ratio of 1.4 with a constant volume
molar specific heat capacity of 208 J.K-1
.mol-1
. The air is at 100 kPa and 17°C at the
beginning of the compression (1 – 2), which decreases its volume by a factor of 8. After
compression, 2320 J heat is isochorically transferred to the gas. Both compression and
expansion processes occur at constant entropy.
a. Using the information provided in the diagram, specify type(s) of thermodynamic
processes associated with the compression and expansion of the air.
b. Determine the temperature, pressure, work, and change in entropy at the end of
each thermodynamic process.

Answer 1

a. The compression and expansion processes shown in the pressure-volume diagram are adiabatic (constant entropy) and isentropic (constant volume) processes, respectively.

b.
Process 1-2: Adiabatic compression
From the pressure-volume diagram, we can see that the compression ratio is 8. Therefore, the final volume at state 2 is 1/8 of the initial volume at state 1:
V2 = V1/8 = 0.02155 m3/mol

Using the ideal gas law:
P1V1 = nRT1
P2V2 = nRT2

Since the amount of air (n = 1.67 mol) is constant throughout the cycle, we can combine these equations to get:
P1V1/T1 = P2V2/T2

Since the compression is adiabatic (Q = 0) and the air is an ideal gas with a constant specific heat ratio, we can use the following relation:

P2/P1 = (V1/V2)γ

where γ = c_P/c_V = 1.4 is the molar specific heat ratio. Substituting the values and solving for P2, we get:

P2 = P1*(V1/V2)γ = 1296.4 kPa

The work done during the compression process is given by:

W12 = -nRT1*(V2 - V1)/(γ - 1)

Substituting the values:

W12 = -1.67*8.31*290*(0.02155 - 0.2155)/(1.4 - 1) = 2212.6 J

The change in entropy during the adiabatic compression is zero:
ΔS12 = 0

From the temperature-entropy diagram, we can see that the temperature at state 2 can be found by following the adiabatic curve that passes through state 1:

T2 = T1*(V1/V2)γ-1 = 773.3 K

Process 2-3: Isochoric heat addition
During this process, heat is transferred to the air at constant volume (isochoric process), so the work done is zero:

W23 = 0

The change in entropy during an isochoric process is given by:

ΔS23 = Q23/T2 = 2320/(1.67*8.31*773.3) = 0.405 J/K

From the ideal gas law:

P3 = P2

T3 = (P3V3)/(nR) = (P2V2)/(nR)

Since the volume is constant during this process, V3 = V2 = 0.02155 m3/mol. Substituting the values, we get:

T3 = (1296.4*0.02155)/(1.67*8.31) = 342.6 K

Process 3-4: Adiabatic expansion
Similar to the adiabatic compression process, we can use the adiabatic relation:

P4/P3 = (V3/V4)γ

where V4 is the final volume at state 4. Since the expansion ratio is the same as the compression ratio, V4 = 8V3 = 0.1724 m3/mol. Substituting the values, we get:

P4 = P3*(V3/V4)γ = 101.2 kPa

The work done during the expansion process is given by:

W34 = -nRT3*(V3 - V4)/(γ - 1)

Substituting the values:

W34 = -1.67*8.31*342.6*(0.1724 - 0.02155)/(1.4 - 1) = -2516.4 J

The change in entropy during the adiabatic expansion is zero:
ΔS34 = 0

From the temperature-entropy diagram, we can see that the temperature at state 4 can be found by following the adiabatic curve that passes through state 3:

T4 = T3*(V3/V4)γ-1 = 200.0 K