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
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)
a. To find the missing P, V, and T values at the end of each process, we can use the ideal gas law equation:
PV = nRT
Where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.
1. Isothermal compression:
Initial Volume (V1) = 9.2 x 10^(-4) m^3
Initial Temperature (T1) = 290 K
Final Volume (V2) = 2.3 x 10^(-4) m^3 (given)
Initial Pressure (P1):
P1 = (n * R * T1) / V1 = (0.183 mol * 8.314 J/(mol·K) * 290 K) / (9.2 x 10^(-4) m^3)
2. Isochoric heating:
Final Volume (V3) = 2.3 x 10^(-4) m^3 (same as previous step)
Final Temperature (T3) = 420 K (given)
Final Pressure (P3):
P3 = (n * R * T3) / V3 = (0.183 mol * 8.314 J/(mol·K) * 420 K) / (2.3 x 10^(-4) m^3)
3. Isobaric expansion:
Final Volume (V4) = 4.4 x 10^(-4) m^3 (given)
Pressure remains constant (P4).
4. Adiabatic cooling:
Initial Volume (V5) = 4.4 x 10^(-4) m^3 (same as previous step)
Initial Temperature (T5) = Temperature at the end of the isobaric expansion (T4, from step 3)
Final Volume (V6) = Initial Volume (V1, from step 1)
Final Temperature (T6) (unknown)
Final Pressure (P6) (unknown)
b. To sketch the corresponding PV diagram, we need to plot the pressure (P) on the y-axis and the volume (V) on the x-axis. We would also indicate the applicable P, V values for each process.
c. To find the corresponding work, internal energy change, and entropy change for each process, we would need additional information such as the nature of the processes (e.g., reversible, irreversible) and any additional data (e.g., heat transfer) provided in the problem statement.
a. To evaluate the missing P, V, T values at the end of each thermodynamic process, we need to apply the appropriate equations for each process.
1. Isothermal compression:
In an isothermal process, the temperature remains constant. The ideal gas law can be used to find the final pressure, P2:
P1V1 = P2V2
Given:
V1 = 9.2 x 10^-4 m^3
T1 = 290 K
V2 = 2.3 x 10^-4 m^3
Using the ideal gas law, we can rearrange it to solve for P2:
P2 = (P1V1)/V2
P2 = (P1 * 9.2 x 10^-4) / (2.3 x 10^-4)
2. Isochoric heating:
In an isochoric process, the volume remains constant. Therefore, the final volume, V2, remains the same. We need to find the final temperature, T2, using the equation:
P1V1/T1 = P2V2/T2
Given:
V1 = 2.3 x 10^-4 m^3
T1 = 290 K
V2 = 2.3 x 10^-4 m^3
T2 = 420 K
We can rearrange the equation to solve for T2:
T2 = (T1 * P2 * V2) / (P1 * V1)
3. Isobaric expansion:
In an isobaric process, the pressure remains constant. Therefore, the final pressure, P2, remains the same. We need to find the final volume, V2, using the equation:
P1V1/T1 = P2V2/T2
Given:
P2 = P2 (same as in the previous step)
T1 = 420 K
P1V1 = P2V2 (from the previous equation)
We can rearrange the equation to solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
4. Adiabatic cooling:
In an adiabatic process, there is no heat exchange with the surroundings. Therefore, the final temperature, T2, remains the same. We need to find the final volume, V2, using the adiabatic equation:
P1 * V1^γ = P2 * V2^γ
Given:
T2 = T1 (same as in the previous step)
P1V1 = P2V2 (from the previous step)
We can rearrange the equation to solve for V2:
V2 = (P1 * V1) / P2
b. To sketch the corresponding PV diagram, we need to plot pressure (P) on the y-axis and volume (V) on the x-axis. We will indicate the applicable P and V values for each process.
c. To find the corresponding work, internal energy change, and entropy change for each process, we need additional information such as the equation of state for an ideal gas, and any other given information about the system (e.g., mass of the gas). Please provide these additional details so that we can continue with the calculations for work, internal energy change, and entropy change.
To answer the given questions, we'll need to apply the laws of thermodynamics and use equations that relate to the different processes involved in the thermodynamic cycle. Here's how we can find the answers step by step:
a) Evaluating the missing P, V, and T values at the end of each process:
1. Isothermal Compression (Process 1)
- We know the initial volume and temperature: V₁ = 9.2 x 10⁻⁴ m³ and T₁ = 290 K.
- Using the equation for an ideal gas: PV = nRT, where P is pressure, n is the number of moles, R is the gas constant, and V is the volume.
- As the process is isothermal, the temperature remains constant throughout. Thus, we can rearrange the equation to solve for the final pressure (P₂):
P₁V₁ = P₂V₂
P₂ = P₁V₁/V₂
Plug in the given values: P₂ = (9.2 x 10⁻⁴ m³) / (2.3 x 10⁻⁴ m³)
2. Isochoric Heating (Process 2)
- This process occurs at constant volume (isochoric), so the volume remains the same (V = 2.3 x 10⁻⁴ m³).
- We need to find the final temperature (T₂). To do this, we use the equation that relates pressure and temperature for an ideal gas at constant volume:
P₁/T₁ = P₂/T₂
T₂ = T₁(P₂/P₁)
Plug in the given values: T₂ = 290 K * (P₂/P₁)
3. Isobaric Expansion (Process 3)
- In this process, the pressure remains constant (P = P₂) throughout.
- We need to find the final volume (V₃). Using the ideal gas equation and rearranging it to solve for V:
P₁V₁ = P₂V₂ = P₃V₃
V₃ = (P₁V₁) / P₂
Plug in the given values: V₃ = (9.2 x 10⁻⁴ m³) / P₂
4. Adiabatic Cooling (Process 4)
- This process is adiabatic, meaning there is no exchange of heat between the system and surroundings.
- To find the final temperature (T₄), we use the adiabatic expansion/compression equation:
T₁V₁^(γ-1) = T₄V₄^(γ-1)
T₄ = T₁(V₁/V₄)^(γ-1)
Since γ for a monatomic gas is 5/3, plug in the given values: T₄ = 290 K * (V₁/V₄)^(2/3)
b) Sketching the corresponding PV diagram:
The PV diagram will have four points corresponding to the initial and final states of each process:
- Point A: Initial state (P₁, V₁)
- Point B: End of isothermal compression (P₂, V₂)
- Point C: End of isochoric heating (P₂, V = 2.3 x 10⁻⁴ m³)
- Point D: End of isobaric expansion (P = P₂, V₃)
- Point E: Final state (P₃, V₄)
Since the values of P, V, and T vary throughout the cycle, plotting the PV diagram will provide a visual representation of the system's behavior.
c) Evaluating the corresponding values for each process:
i. Work (W) for each process can be calculated using the area under the curve on the PV diagram. The work done is given by the equation:
W = -∫PdV
For each process, integrate the area under the curve on the PV diagram to find the work done.
ii. Internal energy change (ΔU) is given by:
ΔU = Q - W
where Q is the heat added or removed during the process and W is the work done.
iii. Entropy change (ΔS) is given by the equation:
ΔS = ∫(dQ/T)
where dQ is the heat added or removed during the process and T is the temperature.
To evaluate the work, internal energy change, and entropy change for each process, we need more information about the specific heat capacity and any heat transfers involved. Please provide any additional information relevant to those calculations.