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Thermodynamics Homework 3

  • Student ID: 1W23CF13
  • Name: TAI Yungche
  • Date: 2025/07/01

Question 1

The initial and final states:

State 1: (V₁, P₁) = (V₀, mP₀)

State 2: (V₂, P₂) = (mV₀, P₀)

Find the slope:

\(a = \frac{P_2 - P_1}{V_2 - V_1} = \frac{P_0 - mP_0}{mV_0 - V_0} = \frac{P_0(1 - m)}{V_0(m - 1)} = -\frac{P_0}{V_0}\)

Using point 1 to find the intercept:

\(P_1 = aV_1 + b\) \(mP_0 = \left(-\frac{P_0}{V_0}\right)V_0 + b\) \(mP_0 = -P_0 + b\) \(b = (m+1)P_0\)

So the pressure function is:

\(P(V) = -\frac{P_0}{V_0}V + (m+1)P_0 = P_0\left(m+1 - \frac{V}{V_0}\right)\)

Question 2

For an ideal gas: \(PV = Nk_BT\), so:

\(T(V) = \frac{P(V)V}{Nk_B}\)

Substituting P(V):

\(T(V) = \frac{V}{Nk_B}\left[P_0\left(m+1 - \frac{V}{V_0}\right)\right] = \frac{P_0}{Nk_B}\left((m+1)V - \frac{V^2}{V_0}\right)\)

This is a quadratic function (downward parabola), so it has a maximum. Taking the derivative:

\(\frac{dT}{dV} = \frac{P_0}{Nk_B}\left((m+1) - \frac{2V}{V_0}\right) = 0\)

Solving for V:

\((m+1) = \frac{2V}{V_0}\)

\(V_{max} = \frac{(m+1)V_0}{2}\)

Since m > 1 (from the diagram), this volume is within the process range [V₀, mV₀].

The maximum temperature is:

\(T_{max} = \frac{P_0}{Nk_B}\left((m+1)V_{max} - \frac{V_{max}^2}{V_0}\right)\)

\(T_{max} = \frac{P_0V_0(m+1)^2}{4Nk_B}\)

Question 3

Process 2 is isobaric compression at constant pressure P₀ from volume mV₀ to V₀.

Work done:

\(W_2 = \int_{mV_0}^{V_0} P_0 dV = P_0(V_0 - mV_0) = -P_0V_0(m-1)\)

For internal energy change:

Initial temperature: \(T_{start} = \frac{P_0(mV_0)}{Nk_B}\)

Final temperature: \(T_{end} = \frac{P_0V_0}{Nk_B}\)

Temperature change: \(\Delta T_2 = \frac{P_0V_0}{Nk_B}(1-m)\)

Internal energy change:

\(\Delta U_2 = C_V \frac{P_0V_0(1-m)}{Nk_B} = -C_V \frac{P_0V_0(m-1)}{Nk_B}\)

From the first law: \(Q_2 = \Delta U_2 + W_2\)

\(Q_2 = -C_V \frac{P_0V_0(m-1)}{Nk_B} - P_0V_0(m-1) = -P_0V_0(m-1)\left(\frac{C_V}{Nk_B} + 1\right)\)

Using \(C_p = C_V + Nk_B\):

\(Q_2 = -P_0V_0(m-1)\frac{C_p}{Nk_B}\)

Question 4

Process 3 is isochoric at constant volume V₀, with pressure changing from P₀ to mP₀.

Work done: \(W_3 = 0\) (no volume change)

Temperature change:

Initial: \(T_{start} = \frac{P_0V_0}{Nk_B}\)

Final: \(T_{end} = \frac{(mP_0)V_0}{Nk_B}\)

Change: \(\Delta T_3 = \frac{P_0V_0(m-1)}{Nk_B}\)

Internal energy change: \(\Delta U_3 = C_V \frac{P_0V_0(m-1)}{Nk_B}\)

\(Q_3 = \Delta U_3 = C_V \frac{P_0V_0(m-1)}{Nk_B}\)

Question 5

For process 1, expansion from (V₀, mP₀) to (mV₀, P₀):

Initial temperature: \(T_1 = \frac{(mP_0)V_0}{Nk_B}\)

Final temperature: \(T_2 = \frac{P_0(mV_0)}{Nk_B}\)

Since T₁ = T₂, the temperature change is zero: \(\Delta T_1 = 0\)

Therefore: \(\Delta U_1 = 0\)

The work done is the area under the P-V curve (trapezoid):

\(W_1 = \frac{1}{2}(mP_0 + P_0)(mV_0 - V_0) = \frac{1}{2}P_0V_0(m+1)(m-1) = \frac{1}{2}P_0V_0(m^2 - 1)\)

Question 6

For the infinitesimal heat exchange: \(dQ = dU + dW\)

Work term: \(dW = P(V)dV = P_0\left(m+1 - \frac{V}{V_0}\right)dV\)

Internal energy term: \(dU = C_VdT\)

From \(T(V) = \frac{P_0}{Nk_B}\left((m+1)V - \frac{V^2}{V_0}\right)\):

\(dT = \frac{P_0}{Nk_B}\left(m+1 - \frac{2V}{V_0}\right)dV\)

So: \(dU = C_V \frac{P_0}{Nk_B}\left(m+1 - \frac{2V}{V_0}\right)dV\)

Combining with \(C_V = 2Nk_B\):

\(dQ = \left[2P_0\left(m+1 - \frac{2V}{V_0}\right) + P_0\left(m+1 - \frac{V}{V_0}\right)\right]dV\)

\(dQ = P_0\left[3(m+1) - \frac{5V}{V_0}\right]dV\)

Heat absorption occurs when dQ > 0, which happens when:

\(3(m+1) - \frac{5V}{V_0} > 0\)

\(V < \frac{3(m+1)V_0}{5}\)

So \(V_a = \frac{3(m+1)V_0}{5}\) is the transition point.

Heat is absorbed from V₀ to V_a, and released from V_a to mV₀.

Question 7

The total heat consumed (positive contributions):

From process 1 (V₀ to V_a):

\(Q_{1,+} = \int_{V_0}^{V_a} P_0\left[3(m+1) - \frac{5V}{V_0}\right]dV\)

After integration and substitution:

\(Q_{1,+} = \frac{P_0V_0}{10}(3m-2)^2\)

From process 2: \(Q_2 < 0\) (heat released)

From process 3: \(Q_3 = 2P_0V_0(m-1) > 0\) (heat absorbed)

Total heat consumed:

\(Q_+ = Q_{1,+} + Q_3 = \frac{P_0V_0}{10}(3m-2)^2 + 2P_0V_0(m-1)\)

\(Q_+ = \frac{P_0V_0}{10}(9m^2 + 8m - 16)\)

Question 8

Total heat released (negative contributions):

From process 1 (V_a to mV₀): \(|Q_{1,-}| = \frac{P_0V_0}{10}(2m-3)^2\)

From process 2: \(|Q_2| = 3P_0V_0(m-1)\)

Total heat released:

\(Q_- = \frac{P_0V_0}{10}(2m-3)^2 + 3P_0V_0(m-1)\)

\(Q_- = \frac{P_0V_0}{10}(4m^2 + 18m - 21)\)

Question 9

The total work for one cycle: \(W = W_1 + W_2 + W_3\)

\(W_1 = \frac{1}{2}P_0V_0(m^2 - 1)\)

\(W_2 = -P_0V_0(m-1)\)

\(W_3 = 0\)

\(W = \frac{1}{2}P_0V_0(m^2 - 1) - P_0V_0(m-1) = \frac{1}{2}P_0V_0(m-1)^2\)

This can also be calculated as the area of the triangle formed by the cycle:

Area = \(\frac{1}{2} \times V_0(m-1) \times P_0(m-1) = \frac{1}{2}P_0V_0(m-1)^2\)

Question 10

The efficiency of the heat engine:

\(\eta = \frac{W}{Q_+} = \frac{\frac{1}{2}P_0V_0(m-1)^2}{\frac{P_0V_0}{10}(9m^2 + 8m - 16)}\)

\(\eta = \frac{5(m-1)^2}{9m^2 + 8m - 16}\)