Thermodynamics Homework 5

• Student ID: 1W23CF13
• Name: TAI Yungche
• Date: 2025-07-20
• Course: Thermodynamics

Question 1

The process 2 is a straight line on the P-V diagram connecting the initial state I at (V₁, P₁) and the final state II at (V₂, P₂). The equation of a straight line is P = mV + c, where m is the slope and c is the y-intercept.

The slope m is given by: \(m = \frac{P_2 - P_1}{V_2 - V_1}\)

The equation for the pressure P as a function of volume V, P(V), is found using the point-slope form of a line:

\(P(V) - P_1 = m (V - V_1)\)

\(P(V) = P_1 + \left(\frac{P_2 - P_1}{V_2 - V_1}\right) (V - V_1)\)

The temperature T as a function of volume V, T(V), is derived from the ideal gas law, \(PV = N k_B T\), where N is the number of molecules and \(k_B\) is the Boltzmann constant.

\(T(V) = \frac{P(V)V}{N k_B}\)

Substituting the expression for P(V):

\(T(V) = \frac{V}{N k_B} \left[ P_1 + \left(\frac{P_2 - P_1}{V_2 - V_1}\right) (V - V_1) \right]\)

Question 2

The differential change in entropy dS for a reversible process in an ideal gas is given by the fundamental thermodynamic relation:

\(T dS = dU + P dV\)

For an ideal gas, the change in internal energy is \(dU = C_V dT\).

\(dS = \frac{C_V}{T} dT + \frac{P}{T} dV\)

To evaluate the entropy change along path 2, this expression can be integrated with respect to V from V₁ to V₂.

\(\Delta_2 S = \int_{V_1}^{V_2} \left( \frac{C_V}{T(V)} \frac{dT}{dV} + \frac{P(V)}{T(V)} \right) dV\)

From the ideal gas law, \(P(V)/T(V) = N k_B / V\).

The derivative \(dT/dV\) is found from the expression for T(V) from Question 1:

\(\frac{dT}{dV} = \frac{d}{dV} \left( \frac{P(V)V}{N k_B} \right) = \frac{1}{N k_B} \left( \frac{dP}{dV}V + P(V) \right)\)

From Question 1, \(\frac{dP}{dV} = m = \frac{P_2 - P_1}{V_2 - V_1}\).

\(\frac{dT}{dV} = \frac{1}{N k_B} (mV + P(V))\)

The first term in the integrand becomes:

\(\frac{C_V}{T(V)} \frac{dT}{dV} = \frac{C_V N k_B}{P(V)V} \frac{1}{N k_B} (mV + P(V)) = C_V \left( \frac{m}{P(V)} + \frac{1}{V} \right)\)

The integrand is then:

\(\frac{dS}{dV} = C_V \left( \frac{m}{P(V)} + \frac{1}{V} \right) + \frac{N k_B}{V} = \frac{C_V + N k_B}{V} + \frac{C_V m}{P(V)}\)

Using the relation for an ideal gas \(C_P = C_V + N k_B\):

\(\frac{dS}{dV} = \frac{C_P}{V} + \frac{C_V m}{P_1 + m(V - V_1)}\)

Now, integrate from V₁ to V₂:

\(\Delta_2 S = \int_{V_1}^{V_2} \frac{C_P}{V} dV + \int_{V_1}^{V_2} \frac{C_V m}{P_1 + m(V - V_1)} dV\)

The first integral is:

\(\int_{V_1}^{V_2} \frac{C_P}{V} dV = C_P [\ln V]_{V_1}^{V_2} = C_P \ln\left(\frac{V_2}{V_1}\right)\)

For the second integral, let \(u = P_1 + m(V-V_1)\), so \(du = m dV\). The limits of integration become \(u(V_1) = P_1\) and \(u(V_2) = P_2\).

\(\int_{P_1}^{P_2} \frac{C_V}{u} du = C_V [\ln u]_{P_1}^{P_2} = C_V \ln\left(\frac{P_2}{P_1}\right)\)

The total entropy change along process 2 is:

\(\Delta_2 S = C_P \ln\left(\frac{V_2}{V_1}\right) + C_V \ln\left(\frac{P_2}{P_1}\right)\)

Using \(C_P = C_V + N k_B\):

\(\Delta_2 S = (C_V + N k_B) \ln\left(\frac{V_2}{V_1}\right) + C_V \ln\left(\frac{P_2}{P_1}\right)\)

\(\Delta_2 S = C_V \left( \ln\left(\frac{V_2}{V_1}\right) + \ln\left(\frac{P_2}{P_1}\right) \right) + N k_B \ln\left(\frac{V_2}{V_1}\right)\)

\(\Delta_2 S = C_V \ln\left(\frac{P_2 V_2}{P_1 V_1}\right) + N k_B \ln\left(\frac{V_2}{V_1}\right)\)

Question 3

The entropy change \(\Delta_1 S\) along process 1 is the sum of the entropy changes for the two stages: the isobaric expansion from I to 1, and the isochoric cooling from 1 to II.

\(\Delta_1 S = \Delta S_{I \to 1} + \Delta S_{1 \to II}\)

Stage 1: Isobaric expansion (I → 1) at constant pressure \(P_1\).

The entropy change for an isobaric process is given by:

\(\Delta S_{I \to 1} = \int_{T_I}^{T_1} \frac{C_P}{T} dT = C_P \ln\left(\frac{T_1}{T_I}\right)\)

From the ideal gas law, \(T_I = P_1 V_1 / (N k_B)\) and \(T_1 = P_1 V_2 / (N k_B)\). Thus, \(T_1/T_I = V_2/V_1\).

\(\Delta S_{I \to 1} = C_P \ln\left(\frac{V_2}{V_1}\right)\)

Stage 2: Isochoric process (1 → II) at constant volume \(V_2\).

The entropy change for an isochoric process is given by:

\(\Delta S_{1 \to II} = \int_{T_1}^{T_{II}} \frac{C_V}{T} dT = C_V \ln\left(\frac{T_{II}}{T_1}\right)\)

From the ideal gas law, \(T_1 = P_1 V_2 / (N k_B)\) and \(T_{II} = P_2 V_2 / (N k_B)\). Thus, \(T_{II}/T_1 = P_2/P_1\).

\(\Delta S_{1 \to II} = C_V \ln\left(\frac{P_2}{P_1}\right)\)

The total entropy change for process 1 is the sum of the changes from the two stages:

\(\Delta_1 S = \Delta S_{I \to 1} + \Delta S_{1 \to II} = C_P \ln\left(\frac{V_2}{V_1}\right) + C_V \ln\left(\frac{P_2}{P_1}\right)\)

This result is identical to the entropy change \(\Delta_2 S\) calculated for process 2, which is expected since entropy is a state function and its change depends only on the initial and final states.

Question 4

The entropy change \(\Delta_3 S\) is evaluated along the isothermal process 3, which occurs at a constant temperature T. For an ideal gas, the differential change in entropy is given by:

\(dS = \frac{dU}{T} + \frac{P}{T} dV\)

The internal energy of an ideal gas depends only on temperature, \(U=U(T)\). In an isothermal process, \(dT=0\), so the change in internal energy \(dU = C_V dT = 0\).

The expression for the entropy change simplifies to:

\(dS = \frac{P}{T} dV\)

Using the ideal gas law, \(PV = N k_B T\), we can write \(P/T = N k_B/V\).

\(dS = N k_B \frac{dV}{V}\)

To find the total entropy change between state I (volume \(V_1\)) and state II (volume \(V_2\)), this expression is integrated:

\(\Delta_3 S = \int_{V_1}^{V_2} N k_B \frac{dV}{V} = N k_B [\ln V]_{V_1}^{V_2}\)

\(\Delta_3 S = N k_B \ln\left(\frac{V_2}{V_1}\right)\)

Question 5

The entropy change \(\Delta_4 S\) along process 4 is the sum of the entropy changes for the two stages: the isochoric cooling from I to 4, and the isobaric expansion from 4 to II.

\(\Delta_4 S = \Delta S_{I \to 4} + \Delta S_{4 \to II}\)

Stage 1: Isochoric process (I → 4) at constant volume \(V_1\).

The entropy change for a constant volume process is:

\(\Delta S_{I \to 4} = \int_{T_I}^{T_4} \frac{C_V}{T} dT = C_V \ln\left(\frac{T_4}{T_I}\right)\)

From the ideal gas law, \(T_I = P_1 V_1 / (N k_B)\) and \(T_4 = P_2 V_1 / (N k_B)\). The ratio of temperatures is \(T_4/T_I = P_2/P_1\).

\(\Delta S_{I \to 4} = C_V \ln\left(\frac{P_2}{P_1}\right)\)

Stage 2: Isobaric process (4 → II) at constant pressure \(P_2\).

The entropy change for a constant pressure process is:

\(\Delta S_{4 \to II} = \int_{T_4}^{T_{II}} \frac{C_P}{T} dT = C_P \ln\left(\frac{T_{II}}{T_4}\right)\)

From the ideal gas law, \(T_4 = P_2 V_1 / (N k_B)\) and \(T_{II} = P_2 V_2 / (N k_B)\). The ratio of temperatures is \(T_{II}/T_4 = V_2/V_1\).

\(\Delta S_{4 \to II} = C_P \ln\left(\frac{V_2}{V_1}\right)\)

The total entropy change for process 4 is the sum:

\(\Delta_4 S = \Delta S_{I \to 4} + \Delta S_{4 \to II} = C_V \ln\left(\frac{P_2}{P_1}\right) + C_P \ln\left(\frac{V_2}{V_1}\right)\)

Question 6

The entropy changes for the four processes are:

\(\Delta_1 S = C_P \ln\left(\frac{V_2}{V_1}\right) + C_V \ln\left(\frac{P_2}{P_1}\right)\)

\(\Delta_2 S = C_P \ln\left(\frac{V_2}{V_1}\right) + C_V \ln\left(\frac{P_2}{P_1}\right)\)

\(\Delta_3 S = N k_B \ln\left(\frac{V_2}{V_1}\right)\)

\(\Delta_4 S = C_P \ln\left(\frac{V_2}{V_1}\right) + C_V \ln\left(\frac{P_2}{P_1}\right)\)

Entropy is a state function, meaning its change depends only on the initial and final states (I and II), not on the path taken. Therefore, the entropy change must be the same for all four processes.

\(\Delta_1 S = \Delta_2 S = \Delta_3 S = \Delta_4 S\)

Equating the expression for \(\Delta_1 S\) and \(\Delta_3 S\):

\(C_P \ln\left(\frac{V_2}{V_1}\right) + C_V \ln\left(\frac{P_2}{P_1}\right) = N k_B \ln\left(\frac{V_2}{V_1}\right)\)

Using the relation for an ideal gas, \(C_P = C_V + N k_B\):

\((C_V + N k_B) \ln\left(\frac{V_2}{V_1}\right) + C_V \ln\left(\frac{P_2}{P_1}\right) = N k_B \ln\left(\frac{V_2}{V_1}\right)\)

\(C_V \ln\left(\frac{V_2}{V_1}\right) + N k_B \ln\left(\frac{V_2}{V_1}\right) + C_V \ln\left(\frac{P_2}{P_1}\right) = N k_B \ln\left(\frac{V_2}{V_1}\right)\)

\(C_V \ln\left(\frac{V_2}{V_1}\right) + C_V \ln\left(\frac{P_2}{P_1}\right) = 0\)

\(C_V \left( \ln\left(\frac{V_2}{V_1}\right) + \ln\left(\frac{P_2}{P_1}\right) \right) = 0\)

\(\ln\left(\frac{P_2 V_2}{P_1 V_1}\right) = 0\)

\(\frac{P_2 V_2}{P_1 V_1} = 1 \implies P_1 V_1 = P_2 V_2\)

This confirms that the initial and final states must lie on the same isotherm, which is consistent with the existence of path 3.

To express the entropy change in terms of \(P_1, P_2, N, k, C_V\), we use the condition \(P_1 V_1 = P_2 V_2\), which implies \(\frac{V_2}{V_1} = \frac{P_1}{P_2}\).

Substituting this into the expression for \(\Delta_3 S\):

\(\Delta_3 S = N k_B \ln\left(\frac{V_2}{V_1}\right) = N k_B \ln\left(\frac{P_1}{P_2}\right)\)

Since all entropy changes are equal, this expression is valid for all four processes.

\(\Delta_1 S = \Delta_2 S = \Delta_3 S = \Delta_4 S = N k_B \ln\left(\frac{P_1}{P_2}\right)\)

Which change is the biggest?

Since all four entropy changes are identical, no single change is the biggest. They are all equal.

Are there identical changes among the four process?

Yes, all four changes in entropy are identical.

Question 1

The given formula for the flux (number of collisions per unit area per unit time) is:

\(\frac{1}{A} \frac{dN}{dt} = n \sqrt{\frac{kT}{2\pi m}}\)

The task is to rewrite this formula in terms of the molar mass \(\mu\) and the universal gas constant \(R\).

The relationships between the microscopic and molar quantities are:

\(R = N_A k \implies k = \frac{R}{N_A}\)

\(\mu = N_A m \implies m = \frac{\mu}{N_A}\)

where \(N_A\) is Avogadro's number.

Substituting these into the term inside the square root:

\(\frac{kT}{m} = \frac{(R/N_A)T}{(\mu/N_A)} = \frac{RT}{\mu}\)

Substituting this back into the original formula gives the rewritten formula:

\(\frac{1}{A} \frac{dN}{dt} = n \sqrt{\frac{RT}{2\pi \mu}}\)

Question 2

The number of collisions per unit time, \(\frac{dN}{dt}\), is the flux multiplied by the area \(A\).

\(\frac{dN}{dt} = A \cdot \left( n \sqrt{\frac{RT}{2\pi \mu}} \right)\)

The concentration \(n\) can be expressed in terms of pressure \(P\) using the ideal gas law \(P = nkT\). Substituting \(k=R/N_A\):

\(P = n \frac{R}{N_A} T \implies n = \frac{P N_A}{RT}\)

Substituting this expression for \(n\) into the formula for the flux:

\(\frac{1}{A} \frac{dN}{dt} = \frac{P N_A}{RT} \sqrt{\frac{RT}{2\pi \mu}} = \frac{P N_A}{\sqrt{2\pi \mu RT}}\)

So, the number of collisions per unit time is:

\(\frac{dN}{dt} = \frac{P A N_A}{\sqrt{2\pi \mu RT}}\)

The given values are:

\(A = 10 \, cm^2 = 10 \times (10^{-2} \, m)^2 = 10^{-3} \, m^2\)

\(\mu = 29 \, g/mol = 0.029 \, kg/mol\)

\(T = 18^\circ C = 18 + 273.15 = 291.15 \, K\)

\(P_0 = 10^5 \, Pa\)

\(R = 8.314 \, J/(mol \cdot K)\)

\(N_A = 6.022 \times 10^{23} \, mol^{-1}\)

First, calculate the denominator:

\(\sqrt{2\pi \mu RT} = \sqrt{2\pi (0.029 \, kg/mol)(8.314 \, J/(mol \cdot K))(291.15 \, K)}\)

\(\sqrt{2\pi \mu RT} = \sqrt{440.55 \, kg^2 \cdot m^2 \cdot s^{-2} \cdot mol^{-2}} = 20.989 \, kg \cdot m \cdot s^{-1} \cdot mol^{-1}\)

Now, calculate the number of collisions per unit time:

\(\frac{dN}{dt} = \frac{(10^5 \, Pa) (10^{-3} \, m^2) (6.022 \times 10^{23} \, mol^{-1})}{20.989 \, kg \cdot m \cdot s^{-1} \cdot mol^{-1}}\)

\(\frac{dN}{dt} = \frac{6.022 \times 10^{25}}{20.989} \, s^{-1} \approx 2.869 \times 10^{24} \, s^{-1}\)

Question 3

The rate of change of the number of molecules \(N\) inside the room is equal to the negative of the flux of molecules escaping through the hole of area \(A\).

\(\frac{dN}{dt} = -A \cdot (\text{flux})\)

The flux is given by the formula from Question 1:

\(\text{flux} = n \sqrt{\frac{RT}{2\pi\mu}}\)

The concentration \(n\) is the number of molecules \(N\) per volume \(V\), so \(n(t) = N(t)/V\). The temperature \(T\) is assumed to be constant.

\(\frac{dN}{dt} = -A \frac{N(t)}{V} \sqrt{\frac{RT}{2\pi\mu}}\)

This is a first-order ordinary differential equation for \(N(t)\).

\(\frac{dN}{N} = -\frac{A}{V} \sqrt{\frac{RT}{2\pi\mu}} dt\)

Let the constant term be denoted by \(1/\tau\):

\(\frac{1}{\tau} = \frac{A}{V} \sqrt{\frac{RT}{2\pi\mu}}\)

The differential equation becomes:

\(\frac{dN}{N} = -\frac{1}{\tau} dt\)

Integrating both sides from \(t=0\) to \(t\):

\(\int_{N(0)}^{N(t)} \frac{dN}{N} = -\int_0^t \frac{1}{\tau} dt'\)

\(\ln\left(\frac{N(t)}{N(0)}\right) = -\frac{t}{\tau}\)

\(N(t) = N(0) e^{-t/\tau}\)

According to the ideal gas law, \(P(t)V = N(t)RT/N_A\). Since \(V\), \(T\), \(R\), and \(N_A\) are constant, pressure is directly proportional to the number of molecules: \(P(t) \propto N(t)\).

Therefore, \(\frac{P(t)}{P(0)} = \frac{N(t)}{N(0)}\).

With \(P(0) = P_0\), the pressure inside the room as a function of time is:

\(P(t) = P_0 e^{-t/\tau} = P_0 \exp\left(-\frac{A t}{V} \sqrt{\frac{RT}{2\pi\mu}}\right)\)

Question 4

The problem asks for the time \(t_{1/2}\) it takes for the pressure to become half of its initial value, i.e., \(P(t_{1/2}) = P_0/2\).

Using the expression for \(P(t)\) from Question 3:

\(\frac{P_0}{2} = P_0 e^{-t_{1/2}/\tau}\)

\(\frac{1}{2} = e^{-t_{1/2}/\tau}\)

Taking the natural logarithm of both sides:

\(\ln\left(\frac{1}{2}\right) = -\frac{t_{1/2}}{\tau}\)

\(-\ln(2) = -\frac{t_{1/2}}{\tau}\)

\(t_{1/2} = \tau \ln(2)\)

where \(\tau = \frac{V}{A \sqrt{RT/(2\pi\mu)}}\).

The given values are:

\(V = 10 \, m^3\)

\(A = 10 \, mm^2 = 10 \times (10^{-3} \, m)^2 = 10^{-5} \, m^2\)

\(\mu = 29 \, g/mol = 0.029 \, kg/mol\)

\(T = 18^\circ C = 291.15 \, K\)

\(R = 8.314 \, J/(mol \cdot K)\)

First, calculate the term in the square root:

\(\sqrt{\frac{RT}{2\pi\mu}} = \sqrt{\frac{(8.314 \, J/(mol \cdot K))(291.15 \, K)}{2\pi(0.029 \, kg/mol)}} = \sqrt{\frac{2420.3}{0.1822}} \, m/s = 115.25 \, m/s\)

Now, calculate the characteristic time \(\tau\):

\(\tau = \frac{10 \, m^3}{(10^{-5} \, m^2)(115.25 \, m/s)} = \frac{10^6}{115.25} \, s = 8676.8 \, s\)

Finally, calculate the half-life \(t_{1/2}\):

\(t_{1/2} = (8676.8 \, s) \times \ln(2) \approx (8676.8 \, s) \times 0.69315 \approx 6013 \, s\)

This is approximately \(100.2\) minutes, or \(1.67\) hours.