Homework 4

• Student ID: 1W23CF13
• Name: TAI Yungche
• Date: 2025-07-11
• Course: Thermodynamics

Question 1

An isothermal process occurs at a fixed temperature, T. The relationship between pressure and volume for the almost ideal gas is given by its equation of state. To find the function P(V) for this process, the equation of state is rearranged.

The given equation of state is:

\(PV = NkT \left(1 + \frac{B}{V}\right)\)

Solving for P as a function of V at constant T:

\(P(V) = \frac{NkT}{V} \left(1 + \frac{B}{V}\right)\)

This can be expanded to:

\(P(V) = \frac{NkT}{V} + \frac{NkTB}{V^2}\)

This equation describes the pressure as a function of volume for an isothermal process of the almost ideal gas.

Question 2

The work W_T, heat Q_T, and change in internal energy ΔU_T for an isothermal expansion from volume \(V_i\) to \(V_f\) at temperature T are determined as follows.

Evaluation of Work Done (W_T) The work done by the gas during an expansion is given by the integral of P dV:

\(W_T = \int_{V_i}^{V_f} P(V) dV\)

Using the expression for P(V) from Question 1:

\(W_T = \int_{V_i}^{V_f} \left(\frac{NkT}{V} + \frac{NkTB}{V^2}\right) dV\)

The integral is evaluated term by term, treating N, k, T, and B as constants:

Evaluation of Change in Internal Energy (ΔU_T) The change in internal energy with volume at constant temperature is given by the thermodynamic identity:

\(\left(\frac{\partial U}{\partial V}\right)_T = T \left(\frac{\partial P}{\partial T}\right)_V - P\)

From the equation of state,

\(P(V, T) = \frac{NkT}{V} + \frac{NkTB}{V^2}\), the partial derivative of P with respect to T at constant V is:

\(\left(\frac{\partial P}{\partial T}\right)_V = \frac{\partial}{\partial T} \left(\frac{NkT}{V} + \frac{NkTB}{V^2}\right) = \frac{Nk}{V} + \frac{NkB}{V^2} = \frac{Nk}{V}\left(1 + \frac{B}{V}\right)\)

Substituting this into the identity:

\(\left(\frac{\partial U}{\partial V}\right)_T = T \left[ \frac{Nk}{V}\left(1 + \frac{B}{V}\right) \right] - P\)

Recognizing that \(P = \frac{NkT}{V}\left(1 + \frac{B}{V}\right)\):

\(\left(\frac{\partial U}{\partial V}\right)_T = P - P = 0\)

This result indicates that the internal energy U depends only on temperature, U = U(T). For an isothermal process, the temperature is constant, so the internal energy does not change.

\(\Delta U_T = 0\)

Evaluation of Heat (Q_T) The heat absorbed by the gas is determined from the First Law of Thermodynamics:

\(\Delta U = Q - W\)

For the isothermal process:

\(\Delta U_T = Q_T - W_T\)

Since \(\Delta U_T = 0\):

\(Q_T = W_T\)

The heat absorbed is equal to the work done by the gas.

\(Q_T = NkT \ln\left(\frac{V_f}{V_i}\right) + NkTB \left(\frac{1}{V_i} - \frac{1}{V_f}\right)\)

Question 3

For an adiabatic process, the first law of thermodynamics states \(dU = -dW\), where \(dQ=0\). The work done by the gas is \(dW = P dV\). From Question 2, the internal energy \(U\) of the almost ideal gas depends only on the temperature \(T\), so \(dU = C_V dT\). The heat capacity \(C_V\) is assumed to be constant. The adiabatic condition is thus \(C_V dT = -P dV\).

The equation of state for the almost ideal gas is

\(PV = NkT \left(1 + \frac{B}{V}\right)\).

To find a differential equation relating \(P\) and \(V\), \(T\) must be eliminated.

From the equation of state,

\(T = \frac{PV}{Nk(1+B/V)} = \frac{PV^2}{Nk(V+B)}\).

The differential \(dT\) is found using the product and quotient rules:

So,

\(dT = \frac{V^2}{Nk(V+B)} dP + \frac{P(V^2+2VB)}{Nk(V+B)^2} dV\).

Substitute \(P\) and \(dT\) into the adiabatic condition \(C_V dT + P dV = 0\):

This equation is of the form

\(\frac{dP}{P} = F(B,V) dV\), where the function \(F(B,V)\) is:

\(F(B,V) = -\left( \frac{V+2B}{V(V+B)} + \frac{Nk}{C_V} \frac{V+B}{V^2} \right)\).

Question 4

The problem is to show that

\(P_A(V) = \frac{\text{const}}{V^\gamma} \frac{B+V}{V} \exp\left(\frac{Nk B}{C_V V}\right)\)

is the solution to the differential equation derived in Question 3, and to find the value of \(\gamma\).

The proposed solution can be written as

\(P_A(V) = K \frac{B+V}{V^{\gamma+1}} \exp\left(\frac{Nk B}{C_V V}\right)\) for some constant \(K\).

Taking the natural logarithm of the solution:

\(\ln P_A(V) = \ln K + \ln(B+V) - (\gamma+1)\ln V + \frac{Nk B}{C_V V}\).

Differentiating with respect to \(V\) to find \(\frac{1}{P_A}\frac{dP_A}{dV}\):

\(\frac{d(\ln P_A)}{dV} = \frac{1}{B+V} - \frac{\gamma+1}{V} - \frac{Nk B}{C_V V^2}\).

From Question 3, the differential equation for an adiabatic process is \(\frac{1}{P}\frac{dP}{dV} = F(B,V)\), where

\(F(B,V) = -\left( \frac{V+2B}{V(V+B)} + \frac{Nk}{C_V} \frac{V+B}{V^2} \right)\).

To compare this with the derivative of the proposed solution, \(F(B,V)\) is simplified using partial fraction decomposition for the first term:

\(\frac{V+2B}{V(V+B)} = \frac{2}{V} - \frac{1}{V+B}\).

And the second term is expanded:

\(\frac{Nk}{C_V}\frac{V+B}{V^2} = \frac{Nk}{C_V}\left(\frac{1}{V} + \frac{B}{V^2}\right)\).

So, \(F(B,V) = -\left( \frac{2}{V} - \frac{1}{V+B} + \frac{Nk}{C_V V} + \frac{Nk B}{C_V V^2} \right) = \frac{1}{V+B} - \frac{1}{V}\left(2+\frac{Nk}{C_V}\right) - \frac{Nk B}{C_V V^2}\).

Equating the two expressions for \(\frac{1}{P}\frac{dP}{dV}\):

\(\frac{1}{B+V} - \frac{\gamma+1}{V} - \frac{Nk B}{C_V V^2} = \frac{1}{V+B} - \frac{1}{V}\left(2+\frac{Nk}{C_V}\right) - \frac{Nk B}{C_V V^2}\).

The terms \(\frac{1}{B+V}\) and \(-\frac{Nk B}{C_V V^2}\) are identical on both sides and cancel out.

\(-\frac{\gamma+1}{V} = -\frac{1}{V}\left(2+\frac{Nk}{C_V}\right)\).

\(\gamma+1 = 2+\frac{Nk}{C_V}\).

\(\gamma = 1+\frac{Nk}{C_V}\).

To express \(\gamma\) in terms of \(C_P, C_V, N, k\), the relation between these quantities is needed. For an almost ideal gas, the leading order approximation for the heat capacities is the same as for an ideal gas, i.e., \(C_P - C_V = Nk\).

Using this, \(\gamma = \frac{C_V+Nk}{C_V} = \frac{C_P}{C_V}\).

Thus, \(\gamma\) is the ratio of the heat capacities.

Question 5

The relation for the Carnot cycle is derived from the adiabatic processes. The relevant equation for an adiabatic process is obtained by integrating

\(C_V d\ln T = -Nk(\frac{1}{V}+\frac{B}{V^2})dV\).

For the adiabatic expansion from b to c (temperature from \(T_1\) to \(T_2\), volume from \(V_b\) to \(V_c\)):

\(C_V \int_{T_1}^{T_2} d\ln T = -Nk \int_{V_b}^{V_c} \left(\frac{1}{V} + \frac{B}{V^2}\right) dV\).

\(C_V (\ln T_2 - \ln T_1) = -Nk \left[ \ln V - \frac{B}{V} \right]_{V_b}^{V_c}\).

\(C_V \ln\left(\frac{T_2}{T_1}\right) = -Nk \left( (\ln V_c - \frac{B}{V_c}) - (\ln V_b - \frac{B}{V_b}) \right)\).

\(-C_V \ln\left(\frac{T_1}{T_2}\right) = -Nk \left( \ln\left(\frac{V_c}{V_b}\right) - \frac{B}{V_c} + \frac{B}{V_b} \right)\).

\(C_V \ln\left(\frac{T_1}{T_2}\right) = Nk \left( \ln\left(\frac{V_c}{V_b}\right) - \frac{B}{V_c} + \frac{B}{V_b} \right)\).

For the adiabatic compression from d to a (temperature from \(T_2\) to \(T_1\), volume from \(V_d\) to \(V_a\)):

\(C_V \int_{T_2}^{T_1} d\ln T = -Nk \int_{V_d}^{V_a} \left(\frac{1}{V} + \frac{B}{V^2}\right) dV\).

\(C_V (\ln T_1 - \ln T_2) = -Nk \left[ \ln V - \frac{B}{V} \right]_{V_d}^{V_a}\).

\(C_V \ln\left(\frac{T_1}{T_2}\right) = -Nk \left( (\ln V_a - \frac{B}{V_a}) - (\ln V_d - \frac{B}{V_d}) \right)\).

\(C_V \ln\left(\frac{T_1}{T_2}\right) = -Nk \left( \ln\left(\frac{V_a}{V_d}\right) - \frac{B}{V_a} + \frac{B}{V_d} \right)\).

Equating the expressions for \(C_V \ln(T_1/T_2)\) from the two adiabatic processes:

\(Nk \left( \ln\left(\frac{V_c}{V_b}\right) - \frac{B}{V_c} + \frac{B}{V_b} \right) = -Nk \left( \ln\left(\frac{V_a}{V_d}\right) - \frac{B}{V_a} + \frac{B}{V_d} \right)\).

\(\ln\left(\frac{V_c}{V_b}\right) - \frac{B}{V_c} + \frac{B}{V_b} = -\ln\left(\frac{V_a}{V_d}\right) + \frac{B}{V_a} - \frac{B}{V_d}\).

Using \(\ln(x/y) = -\ln(y/x)\):

\(-\ln\left(\frac{V_b}{V_c}\right) - \frac{B}{V_c} + \frac{B}{V_b} = \ln\left(\frac{V_d}{V_a}\right) + \frac{B}{V_a} - \frac{B}{V_d}\).

This expression is not directly the one in the problem. Let us define \(X_{ij} = \ln(V_i/V_j) - B/V_i + B/V_j\). The problem asks to prove \(X_{bc} = X_{ad}\).

The term from the b->c process is \(Y_{bc} = \ln(V_c/V_b) - B/V_c + B/V_b\). Note that \(Y_{bc} = -\ln(V_b/V_c) - B/V_c + B/V_b\).

It can be shown that \(Y_{bc} = -X_{bc}\), since \(X_{bc} + Y_{bc} = (\ln(V_b/V_c)-\ln(V_b/V_c)) + (-B/V_b+B/V_b) + (B/V_c-B/V_c) = 0\).

The equality derived from the two processes is \(Y_{bc} = -X_{ad}\).

Substituting \(Y_{bc} = -X_{bc}\), we get \(-X_{bc} = -X_{ad}\), which implies \(X_{bc} = X_{ad}\).

Thus, the relation holds:

\(\ln\left(\frac{V_b}{V_c}\right) - \frac{B}{V_b} + \frac{B}{V_c} = \ln\left(\frac{V_a}{V_d}\right) - \frac{B}{V_a} + \frac{B}{V_d}\).

To find \(\alpha\), we use the relation for one of the adiabatic paths, e.g., d->a:

\(C_V \ln\left(\frac{T_1}{T_2}\right) = -Nk X_{ad}\).

Since \(X_{ad} = X_{bc}\),

\(C_V \ln\left(\frac{T_1}{T_2}\right) = -Nk X_{bc}\).

\(X_{bc} = -\frac{C_V}{Nk} \ln\left(\frac{T_1}{T_2}\right)\).

Comparing this with the given equation \(\alpha \ln(T_1/T_2) = X_{bc}\), we find: \(\alpha = -\frac{C_V}{Nk}\).

Question 6

The total heat absorbed, \(Q_1\), occurs during the isothermal expansion from state a to state b at temperature \(T_1\). The total heat released, \(Q_2\), occurs during the isothermal compression from state c to state d at temperature \(T_2\).

From the result of Question 2, the heat exchanged during an isothermal process from volume \(V_i\) to \(V_f\) at temperature \(T\) is:

\(Q_T = NkT \ln\left(\frac{V_f}{V_i}\right) + NkTB \left(\frac{1}{V_i} - \frac{1}{V_f}\right)\).

For the isothermal expansion a → b at \(T_1\):

The heat absorbed is \(Q_1\). The initial state is \((V_a, T_1)\) and the final state is \((V_b, T_1)\).

\(Q_1 = NkT_1 \ln\left(\frac{V_b}{V_a}\right) + NkT_1B \left(\frac{1}{V_a} - \frac{1}{V_b}\right)\).

For the isothermal compression c → d at \(T_2\):

The heat exchanged is \(Q_{cd}\). The initial state is \((V_c, T_2)\) and the final state is \((V_d, T_2)\).

\(Q_{cd} = NkT_2 \ln\left(\frac{V_d}{V_c}\right) + NkT_2B \left(\frac{1}{V_c} - \frac{1}{V_d}\right)\).

Since \(V_d < V_c\), \(Q_{cd}\) is negative, indicating heat is released. The heat released \(Q_2\) is defined as a positive quantity, so \(Q_2 = -Q_{cd}\).

\(Q_2 = - \left( NkT_2 \ln\left(\frac{V_d}{V_c}\right) + NkT_2B \left(\frac{1}{V_c} - \frac{1}{V_d}\right) \right)\)

\(Q_2 = NkT_2 \ln\left(\frac{V_c}{V_d}\right) + NkT_2B \left(\frac{1}{V_d} - \frac{1}{V_c}\right)\).

Question 7

To show that \(\frac{Q_1}{T_1} = \frac{Q_2}{T_2}\), we use the expressions for \(Q_1\) and \(Q_2\) from Question 6.

\(\frac{Q_1}{T_1} = Nk \left[ \ln\left(\frac{V_b}{V_a}\right) + B \left(\frac{1}{V_a} - \frac{1}{V_b}\right) \right]\)

\(\frac{Q_2}{T_2} = Nk \left[ \ln\left(\frac{V_c}{V_d}\right) + B \left(\frac{1}{V_d} - \frac{1}{V_c}\right) \right]\)

The adiabatic relation connects the states at the ends of the adiabatic processes. For an adiabatic process, \(C_V d\ln T = -Nk(\frac{1}{V}+\frac{B}{V^2})dV\).

Integrating for the process b → c (from \(T_1, V_b\) to \(T_2, V_c\)):

\(C_V \ln\left(\frac{T_2}{T_1}\right) = -Nk \left[ \ln(V) - \frac{B}{V} \right]_{V_b}^{V_c} = -Nk \left( \ln\frac{V_c}{V_b} - \frac{B}{V_c} + \frac{B}{V_b} \right)\).

Integrating for the process d → a (from \(T_2, V_d\) to \(T_1, V_a\)):

\(C_V \ln\left(\frac{T_1}{T_2}\right) = -Nk \left[ \ln(V) - \frac{B}{V} \right]_{V_d}^{V_a} = -Nk \left( \ln\frac{V_a}{V_d} - \frac{B}{V_a} + \frac{B}{V_d} \right)\).

From the first adiabatic process, we can write:

\(C_V \ln\left(\frac{T_1}{T_2}\right) = Nk \left( \ln\frac{V_c}{V_b} - \frac{B}{V_c} + \frac{B}{V_b} \right)\).

Equating the two expressions for \(C_V \ln(T_1/T_2)\):

\(Nk \left( \ln\frac{V_c}{V_b} - \frac{B}{V_c} + \frac{B}{V_b} \right) = -Nk \left( \ln\frac{V_a}{V_d} - \frac{B}{V_a} + \frac{B}{V_d} \right)\).

\(\ln V_c - \ln V_b - \frac{B}{V_c} + \frac{B}{V_b} = -(\ln V_a - \ln V_d - \frac{B}{V_a} + \frac{B}{V_d})\).

\(\ln V_c - \ln V_b - \frac{B}{V_c} + \frac{B}{V_b} = -\ln V_a + \ln V_d + \frac{B}{V_a} - \frac{B}{V_d}\).

Rearranging the terms to group those related to the isothermal processes:

\(\ln V_b - \ln V_a + \frac{B}{V_a} - \frac{B}{V_b} = \ln V_c - \ln V_d + \frac{B}{V_d} - \frac{B}{V_c}\).

\(\ln\left(\frac{V_b}{V_a}\right) + B\left(\frac{1}{V_a} - \frac{1}{V_b}\right) = \ln\left(\frac{V_c}{V_d}\right) + B\left(\frac{1}{V_d} - \frac{1}{V_c}\right)\).

Multiplying both sides by \(Nk\):

\(Nk\left[\ln\left(\frac{V_b}{V_a}\right) + B\left(\frac{1}{V_a} - \frac{1}{V_b}\right)\right] = Nk\left[\ln\left(\frac{V_c}{V_d}\right) + B\left(\frac{1}{V_d} - \frac{1}{V_c}\right)\right]\).

This shows that \(\frac{Q_1}{T_1} = \frac{Q_2}{T_2}\).

Question 8

The efficiency \(\eta\) of a heat engine is defined as the net work done \(W\) divided by the heat absorbed from the hot reservoir, \(Q_1\).

\(\eta = \frac{W}{Q_1}\)

For a cyclic process, the net work done is equal to the net heat absorbed.

\(W = Q_{in} - Q_{out} = Q_1 - Q_2\).

So, the efficiency is:

\(\eta = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1}\).

From the result of Question 7, we have the relation \(\frac{Q_1}{T_1} = \frac{Q_2}{T_2}\). This can be rearranged to find the ratio \(\frac{Q_2}{Q_1}\):

\(\frac{Q_2}{Q_1} = \frac{T_2}{T_1}\).

Substituting this ratio into the expression for efficiency:

\(\eta = 1 - \frac{T_2}{T_1}\).