An ideal gas for which Cv = 3R/2 is the working substance of a Carnot engine. During the isothermal expansion, the volume doubles. The ratio of the final volume to the initial volume in the adiabatic expansion is 5.7. The work output of the engine is 9 x 105 J in each cycle. Compute the temperature of the reservoirs between which the engine operates
To solve this problem, we will use the Carnot efficiency formula:
η = 1 - (Tc/Th)
Where η is the Carnot efficiency, Tc is the temperature of the cold reservoir, and Th is the temperature of the hot reservoir.
We can also calculate the work output of the Carnot engine using the formula:
W = Qh - Qc
where W is the work output, Qh is the heat absorbed from the hot reservoir, and Qc is the heat released to the cold reservoir.
Given that Qh = Qc, we can rewrite the equation as:
W = Qh - Qh = 0
Since the work output is zero, the heat absorbed and released must be equal in each cycle.
Now let's solve the problem step by step:
Step 1: Calculate the isothermal expansion work
We know that the work done during the isothermal expansion is given by:
W_expansion = nRT ln(Vf/Vi)
Where n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, Vi is the initial volume, and Vf is the final volume.
In this case, the volume doubles during the isothermal expansion, so Vf = 2Vi.
Therefore,
W_expansion = nRT ln(2)
Step 2: Calculate the adiabatic expansion work
We know that the ratio of the final volume to the initial volume in the adiabatic expansion is 5.7. This implies that the final volume is 5.7 times the initial volume, so Vf = 5.7Vi.
The work done during the adiabatic expansion is given by:
W_adiabatic = (Cv - Cp) ΔT
Where Cp is the specific heat capacity at constant pressure and ΔT is the change in temperature.
Since Cp = Cv + R, we can rewrite the equation as:
W_adiabatic = (Cv - (Cv + R)) ΔT = -R ΔT
Since the adiabatic expansion is an isentropic (no heat transfer) process, ΔT = Tc - Th.
Step 3: Calculate the Carnot efficiency
Since the work output is zero (W = 0), we have:
0 = Qh - Qc = Qh - Qh = 0
Therefore, the heat absorbed from the hot reservoir is equal to the heat released to the cold reservoir.
From the first law of thermodynamics, we know that the net heat transfer during the process is equal to the change in internal energy:
Qh - Qc = ΔU
Since the process is reversible and the ideal gas is an ideal gas, ΔU = 0.
So we have:
0 = Qh - Qc
Using the heat capacities relation Cp - Cv = R, we can rewrite the equation as:
0 = Qh - Qc = (Cp - R) ΔTc - (Cp - R) ΔTh
Dividing both sides of the equation by (Cp - R), we get:
0 = ΔTc - ΔTh
Therefore, the temperature change between the hot reservoir and the cold reservoir is equal.
Step 4: Calculate the temperature of the reservoirs
Since the temperature change between the hot reservoir and the cold reservoir is equal, we have:
ΔTc = ΔTh
Let's define the initial temperature as To, and the temperature after the isothermal expansion as Th.
Since the volume doubles during the isothermal expansion, we have:
Vf = 2Vi
Using the ideal gas law, we can write:
(Pf Vi) / Th = (Po Vi) / To
Where Pf is the pressure after the isothermal expansion, Po is the initial pressure, and Vi is the initial volume.
We can solve for Th:
Th = To * (Pf / Po)
From the ideal gas law, we know that PV = nRT. Since the amount of gas is constant throughout the process, we have:
Po * Vi = Pf * Vf
Substituting Vf = 2Vi, we get:
Po * Vi = Pf * 2Vi
Therefore, Pf = Po / 2.
Substituting this in the equation for Th, we have:
Th = To * (0.5)
Therefore, we can conclude that the temperature of the hot reservoir is half of the initial temperature.
Since the temperature change between the hot reservoir and the cold reservoir is equal, we have:
ΔTc = ΔTh = To - Tc
Therefore, Tc = To - ΔTc.
Substituting ΔTc = To - 0.5To = 0.5To, we get:
Tc = 0.5To
Step 5: Calculate the Carnot efficiency
Using the Carnot efficiency formula:
η = 1 - (Tc/Th)
We can substitute Tc = 0.5To and Th = 0.5To, and simplify:
η = 1 - (0.5To / 0.5To) = 1 - 1 = 0
Since the Carnot efficiency is zero, we have:
η = W/Qh
0 = 9 x 10^5 J / Qh
Therefore, Qh = infinity.
Step 6: Conclusion
From the above calculations, we can conclude that the temperature of the hot reservoir is half of the initial temperature (To/2), and the temperature of the cold reservoir is half of the hot reservoir temperature (To/4).
To solve this problem, we can use the principles of the Carnot cycle and the properties of the ideal gas to find the temperatures of the reservoirs.
The Carnot engine operates on a reversible cycle, which consists of four processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
Let's denote the initial temperature and volume of the gas as T1 and V1, respectively. The final temperature and volume after the isothermal expansion are denoted as T2 and V2, respectively.
First, let's calculate T2. Since the expansion is isothermal, we can use the ideal gas law:
PV = nRT
For an ideal gas undergoing isothermal expansion, the pressure-volume product remains constant (P1V1 = P2V2). Therefore:
P1V1 = P2V2
Since the volume doubles during expansion (V2 = 2V1), we can rewrite the equation as:
P1V1 = P2(2V1) -----> P2 = P1/2
Since we know that Cv = 3R/2, we can use the equation for the molar heat capacity at constant volume (Cv) to relate the initial and final temperatures:
Cv = (R / (γ - 1))
Where γ is the ratio of specific heats. For an ideal monoatomic gas, γ = 5/3.
Cv = (R / (5/3 - 1)) = (3R / 2)
Therefore, we have:
(3R / 2) = (R / (γ - 1))
Solving for γ, we find:
(3/2) = (1 / (γ - 1)) -----> γ - 1 = 2/3 -----> γ = 5/3
Now, we can use the relationship between temperatures during isothermal expansion:
T2 / T1 = (V1 / V2)
Substituting the known values:
T2 / T1 = (V1 / 2V1) = 1/2
Therefore, T2 = (1/2)T1
Next, let's calculate the final temperature after the adiabatic expansion. During adiabatic expansion, the ratio of pressure to volume raised to the power of γ should remain constant:
P2V2^γ = P3V3^γ
Since the volume ratio is given as 5.7 (V3 / V2 = 5.7), we can use this information to solve for P3:
(P2 / P3) = (V3 / V2)^γ = (5.7)^γ
Now, we can use the work done during the adiabatic expansion to find P2:
W_ad = (Cv / (γ - 1)) * (T3 - T2)
Since the work output is given as 9 x 10^5 J and Cv = (3R / 2), we have:
9 x 10^5 J = ((3R / 2) / (γ - 1)) * (T3 - T2)
Substituting γ = 5/3 and T2 = (1/2)T1, we can simplify the equation:
9 x 10^5 J = ((3R / 2) / (5/3 - 1)) * (T3 - (1/2)T1)
Simplifying further, we get:
9 x 10^5 J = (3R / (2/3)) * (T3 - (1/2)T1)
9 x 10^5 J = (9R / 2) * (T3 - (1/2)T1)
Now, we can rearrange the equation to solve for T3 - (1/2)T1:
T3 - (1/2)T1 = (9 x 10^5 J / (9R / 2)) = (2 x 10^5 J) / (R / 2) = 4 x 10^5 J * (2 / R) = 8 x 10^5 J / R
Finally, we can substitute T2 = (1/2)T1 and T3 - (1/2)T1 = 8 x 10^5 J / R into the equation:
T2 / T1 = (V1 / V2) -----> (1/2)T1 / T1 = (V1 / 2V1)
This simplifies to:
1/2 = 1/2
Since this equation is true, we can conclude that the temperature ratio is satisfied. This means that the temperature of the reservoirs between which the engine operates is:
T1 / T3 = 1/2
Therefore, the temperature of the reservoirs is in the ratio 1:2, with T1 being the lower temperature reservoir.