Answer
We can use the work-energy principle to find the net work done on the gas during the cycle. The work done on the gas during an adiabatic process is given by:
W = (P1V1 - P2V2)/(γ - 1)
where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and γ is the adiabatic index of the gas. Since the gas is compressed adiabatically to half of its original volume, we have:
V2 = V1/2
Let's denote the pressure of the gas after compression as P3. Using the ideal gas law, we can relate the pressure, volume, and temperature of the gas before and after compression:
P1V1/T1 = P3V2/T1
Since the temperature of the gas remains constant during the entire cycle, we can simplify this expression as:
P1V1 = P3V2
Substituting V2 = V1/2, we get:
P1 = 2P3
Using the same equation for work done during adiabatic compression, we can relate P1 and P3 as:
W1 = (P1V1 - P3V2)/(γ - 1)
Substituting the above value of P1 and V2 = V1/2, we get:
W1 = (3/2)(P3V1)/(γ - 1)
Next, the gas is cooled at constant volume back to its original temperature of 288 K. Since the volume remains constant during this process, the work done on the gas is zero.
Finally, the gas is allowed to expand isothermally back to its original volume at a constant temperature of 288 K. Using the ideal gas law again, we can relate the pressure and volume of the gas before and after expansion:
P3V2 = P4V1
Since the volume and temperature remain constant during this process, the pressure of the gas after expansion is:
P4 = P3(V2/V1) = P3/2
Using the same equation for work done during isothermal expansion, we can relate P3 and P4 as:
W2 = (P3V1/2)(ln(2))
Substituting the given value for the net work done on the gas during the cycle, we get:
W1 + W2 = (3/2)(P3V1)/(γ - 1) + (P3V1/2)(ln(2)) = 436 J
Substituting γ = 1.40, V1 = 25.0 L, and solving for P3, we get:
P3 ≈ 256 kPa
Therefore, the pressure of the gas at the start of the cycle was approximately 2P3, or 512 kP
Work Step by Step
We can use the work-energy principle to find the net work done on the gas during the cycle. The work done on the gas during an adiabatic process is given by:
W = (P1V1 - P2V2)/(γ - 1)
where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and γ is the adiabatic index of the gas. Since the gas is compressed adiabatically to half of its original volume, we have:
V2 = V1/2
Let's denote the pressure of the gas after compression as P3. Using the ideal gas law, we can relate the pressure, volume, and temperature of the gas before and after compression:
P1V1/T1 = P3V2/T1
Since the temperature of the gas remains constant during the entire cycle, we can simplify this expression as:
P1V1 = P3V2
Substituting V2 = V1/2, we get:
P1 = 2P3
Using the same equation for work done during adiabatic compression, we can relate P1 and P3 as:
W1 = (P1V1 - P3V2)/(γ - 1)
Substituting the above value of P1 and V2 = V1/2, we get:
W1 = (3/2)(P3V1)/(γ - 1)
Next, the gas is cooled at constant volume back to its original temperature of 288 K. Since the volume remains constant during this process, the work done on the gas is zero.
Finally, the gas is allowed to expand isothermally back to its original volume at a constant temperature of 288 K. Using the ideal gas law again, we can relate the pressure and volume of the gas before and after expansion:
P3V2 = P4V1
Since the volume and temperature remain constant during this process, the pressure of the gas after expansion is:
P4 = P3(V2/V1) = P3/2
Using the same equation for work done during isothermal expansion, we can relate P3 and P4 as:
W2 = (P3V1/2)(ln(2))
Substituting the given value for the net work done on the gas during the cycle, we get:
W1 + W2 = (3/2)(P3V1)/(γ - 1) + (P3V1/2)(ln(2)) = 436 J
Substituting γ = 1.40, V1 = 25.0 L, and solving for P3, we get:
P3 ≈ 256 kPa
Therefore, the pressure of the gas at the start of the cycle was approximately 2P3, or 512 kP
