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Net Work Done in Thermodynamic Gas Process

Created by @nathanedwards
 at November 1st 2023, 1:54:07 am.
AP_Physics_2-Questions
#thermodynamics
#work-done
#adiabatic-process

Question:

A gas undergoes a thermodynamic process described by the following steps:

  1. The gas is compressed adiabatically and reversibly from an initial volume of V1V_1 to a final volume of V2V_2. The pressure of the gas doubles during this compression.
  2. The gas is kept at constant pressure and heat is added to it until its temperature doubles.
  3. The gas is allowed to expand isothermally to its initial volume.

Assuming the gas behaves ideally, determine the net work done by the gas during the entire process in terms of P1P_1 (the initial pressure of the gas), V1V_1, and RR (the ideal gas constant).

Answer:

To find the net work done by the gas during the entire process, we need to consider the work done in each step separately and sum them up.

Step 1: Adiabatic and reversible compression

In an adiabatic process, there is no heat exchange with the surroundings (Q=0Q = 0). For an ideal gas, the adiabatic process is described by the equation:

PVγ=constantPV^\gamma = \text{constant}

where PP is the pressure, VV is the volume, and γ=CpCv\gamma = \frac{C_p}{C_v} is the heat capacity ratio. Since the pressure doubles during the compression, we can write:

P2V2γ=2P1V1γP_2V_2^\gamma = 2P_1V_1^\gamma

Since the compression is reversible, we can use the ideal gas law:

PV=nRTPV = nRT

where nn is the number of moles of gas and TT is the temperature. From this equation, we can express the initial and final temperatures in terms of the initial and final volumes:

T1=P1V1nRandT2=2P1V1nRT_1 = \frac{P_1V_1}{nR} \quad \text{and} \quad T_2 = \frac{2P_1V_1}{nR}

Using the adiabatic process equation, we can express the final volume in terms of the initial volume:

V2=(2V1γ)1γ=21γV1V_2 = \left(2V_1^\gamma\right)^{\frac{1}{\gamma}} = 2^{\frac{1}{\gamma}}V_1

Substituting these values into the ideal gas law, we can find the initial and final pressures:

P1=nRT1V1andP2=nRT2V2P_1 = \frac{nRT_1}{V_1} \quad \text{and} \quad P_2 = \frac{nRT_2}{V_2}

Substituting these pressures into the equation P2=2P1P_2 = 2P_1, we can solve for nn:

2(nRT1V1)=nRT2V22\left(\frac{nRT_1}{V_1}\right) = \frac{nRT_2}{V_2}

Simplifying and canceling terms:

2T1V2=T2V12T_1V_2 = T_2V_1

Substituting the expressions for T1T_1, T2T_2, and V2V_2:

2(P1V1nR)(21γV1)=2P1V1nR2\left(\frac{P_1V_1}{nR}\right)\left(2^{\frac{1}{\gamma}}V_1\right) = \frac{2P_1V_1}{nR}

Cancelling out P1V1P_1V_1 and rearranging:

21γ=22^{\frac{1}{\gamma}} = 2

Since the equation is satisfied, we know that the given compression is adiabatic and reversible.

The work done in an adiabatic process can be calculated using the equation:

W=1γ−1(P2V2−P1V1)W = \frac{1}{\gamma - 1}(P_2V_2 - P_1V_1)

Substituting the known values:

W1=1γ−1(2P1V1⋅21γV1−P1V1)W_1 = \frac{1}{\gamma - 1}(2P_1V_1 \cdot 2^{\frac{1}{\gamma}}V_1 - P_1V_1)

Simplifying:

W1=1γ−1(2P1⋅21γV12−P1V1)W_1 = \frac{1}{\gamma - 1}(2P_1 \cdot 2^{\frac{1}{\gamma}}V_1^2 - P_1V_1)
W1=P1V1(1γ−1(2⋅21γ−1))W_1 = P_1V_1 \left(\frac{1}{\gamma - 1}(2 \cdot 2^{\frac{1}{\gamma}} - 1)\right)

Step 2: Constant pressure and heat addition

In this step, the gas is kept at constant pressure (P=P2P = P_2) and heat is added until its temperature doubles (T=2T1T = 2T_1). The work done in this step is given by:

W2=P2(V2−V1)W_2 = P_2(V_2 - V_1)

Substituting the known values:

W2=(2P1)(21γV1−V1)W_2 = (2P_1)(2^{\frac{1}{\gamma}}V_1 - V_1)

Simplifying:

W2=2P1(21γ−1)V1W_2 = 2P_1 \left(2^{\frac{1}{\gamma}} - 1\right)V_1

Step 3: Isothermal expansion

In this step, the gas undergoes an isothermal expansion from V2V_2 to V1V_1. The work done in an isothermal expansion is given by:

W3=nRT1ln⁡(V1V2)W_3 = nRT_1\ln\left(\frac{V_1}{V_2}\right)

Substituting the known values:

W3=nRT1nRln⁡(V121γV1)W_3 = \frac{nRT_1}{nR} \ln\left(\frac{V_1}{2^{\frac{1}{\gamma}}V_1}\right)

Simplifying:

W3=T1ln⁡(121γ)V1W_3 = T_1 \ln\left(\frac{1}{2^{\frac{1}{\gamma}}}\right)V_1

The net work done by the gas during the entire process is given by the sum of the work done in each step:

Wnet=W1+W2+W3W_{\text{net}} = W_1 + W_2 + W_3

Substituting the expressions for W1W_1, W2W_2, and W3W_3:

Wnet=P1V1(1γ−1(2⋅21γ−1))+2P1(21γ−1)V1+T1ln⁡(121γ)V1W_{\text{net}} = P_1V_1 \left(\frac{1}{\gamma - 1}(2 \cdot 2^{\frac{1}{\gamma}} - 1)\right) + 2P_1 \left(2^{\frac{1}{\gamma}} - 1\right)V_1 + T_1 \ln\left(\frac{1}{2^{\frac{1}{\gamma}}}\right)V_1

Using logarithm properties to simplify:

Wnet=P1V1(1γ−1(2⋅21γ−1)+21γ−1−ln⁡(21γ))W_{\text{net}} = P_1V_1 \left(\frac{1}{\gamma - 1}(2 \cdot 2^{\frac{1}{\gamma}} - 1) + 2^{\frac{1}{\gamma}} - 1 - \ln(2^{\frac{1}{\gamma}})\right)
Wnet=P1V1(1γ−1(2⋅21γ−1+(γ−1)(21γ−1))−ln⁡(21γ))W_{\text{net}} = P_1V_1 \left(\frac{1}{\gamma - 1}\left(2 \cdot 2^{\frac{1}{\gamma}} - 1 + (\gamma - 1)(2^{\frac{1}{\gamma}} - 1)\right) - \ln(2^{\frac{1}{\gamma}})\right)
Wnet=P1V1(2⋅21γ−1γ−1−ln⁡(21γ))W_{\text{net}} = P_1V_1 \left(\frac{2\cdot 2^{\frac{1}{\gamma}} - 1}{\gamma - 1} - \ln(2^{\frac{1}{\gamma}})\right)

Simplifying further:

Wnet=P1V1(21γ(2−1)−1γ−1−ln⁡(21γ))W_{\text{net}} = P_1V_1 \left(\frac{2^{\frac{1}{\gamma}}(2 - 1) - 1}{\gamma - 1} - \ln(2^{\frac{1}{\gamma}})\right)
Wnet=P1V1(21γ−1γ−1−ln⁡(21γ))W_{\text{net}} = P_1V_1 \left(\frac{2^{\frac{1}{\gamma}} - 1}{\gamma - 1} - \ln(2^{\frac{1}{\gamma}})\right)

Therefore, the net work done by the gas during the entire process is equal to P1V1(21γ−1γ−1−ln⁡(21γ))P_1V_1 \left(\frac{2^{\frac{1}{\gamma}} - 1}{\gamma - 1} - \ln(2^{\frac{1}{\gamma}})\right).

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