Homework 1

• Student ID: 1W23CF13
• Name: TAI Yungche
• Date: 2025/06/22

Water Vapour

Question 1

The mass of a single water molecule can be calculated from the molar mass and Avogadro's number:

\[m = \frac{\mu}{N_A} = \frac{18 \text{ g/mol}}{6.022 \times 10^{23} \text{ molecules/mol}}\]

Converting to kg:

\[m = \frac{18 \times 10^{-3} \text{ kg/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} = 2.99 \times 10^{-26} \text{ kg}\]

Question 2

The thermal velocity is given by:

\[v_T = \sqrt{\langle v^2 \rangle}\]

For an ideal gas, the mean square velocity is related to temperature through the equipartition theorem:

\[\frac{1}{2}m\langle v^2 \rangle = \frac{3}{2}k_B T\]

where \(k_B = 1.38 \times 10^{-23}\) J/K is Boltzmann's constant.

Therefore:

\[\langle v^2 \rangle = \frac{3k_B T}{m}\]

At \(T = 15°C = 288.15\) K:

\[\langle v^2 \rangle = \frac{3 \times 1.38 \times 10^{-23} \times 288.15}{2.99 \times 10^{-26}} = 3.99 \times 10^{5} \text{ m}^2/\text{s}^2\]

\[v_T = \sqrt{3.99 \times 10^{5}} = 632 \text{ m/s}\]

Question 3

First, calculate the mass of water:

\[m_{H_2O} = \rho_{H_2O} \times V_{H_2O} = 1000 \text{ kg/m}^3 \times 150 \times 10^{-6} \text{ m}^3 = 0.15 \text{ kg}\]

The number of water molecules in the bottle:

\[N = \frac{m_{H_2O}}{m} = \frac{0.15}{2.99 \times 10^{-26}} = 5.02 \times 10^{24} \text{ molecules}\]

When all water evaporates and distributes uniformly in the room, the concentration of water molecules in the air:

\[n = \frac{N}{V_{Room}} = \frac{5.02 \times 10^{24}}{20} = 2.51 \times 10^{23} \text{ molecules/m}^3\]

Question 4

For an ideal gas, the pressure is given by:

\[P = nk_BT\]

where \(n\) is the number density of molecules.

\[P_{H_2O} = 2.51 \times 10^{23} \times 1.38 \times 10^{-23} \times 288.15\]

\[P_{H_2O} = 998 \text{ Pa}\]

Comparing to atmospheric pressure:

\[\frac{P_{H_2O}}{P_0} = \frac{998}{1 \times 10^5} = 0.00998 \approx 0.01\]

The water vapor pressure is about 1% of atmospheric pressure.

Question 5

The relative humidity is:

\[\phi = \frac{P_{H_2O}}{P_s} \times 100\%\]

\[\phi = \frac{998}{1.705 \times 10^3} \times 100\%\]

\[\phi = 58.5\%\]

Question 6

The total kinetic energy of all water molecules:

\[U = \frac{3}{2}Nk_BT\]

\[U = \frac{3}{2} \times 5.02 \times 10^{24} \times 1.38 \times 10^{-23} \times 288.15\]

\[U = 2.99 \times 10^4 \text{ J}\]

Question 7

Converting 1 kWh to SI units:

\[1 \text{ kWh} = 1 \text{ kW} \times 1 \text{ h} = 1000 \text{ W} \times 3600 \text{ s}\]

\[1 \text{ kWh} = 3.6 \times 10^6 \text{ J}\]

Expressing the internal energy in kWh:

\[U = \frac{2.99 \times 10^4}{3.6 \times 10^6} = 0.00831 \text{ kWh}\]

Relativistic Ideal Gas

Question 1

For a general momentum-velocity relation \(\vec{p} = f(v^2)\vec{v}\), the pressure formula needs to account for the momentum transfer at the walls.

When a molecule with momentum \(p_x\) collides elastically with a wall perpendicular to the x-axis, the momentum change is \(\Delta p_x = 2p_x\).

The rate of collisions with one wall for molecules with velocity \(v_x > 0\) is: $\(\frac{N}{V} \cdot v_x \cdot \frac{1}{2}\)$

where the factor 1/2 accounts for molecules moving toward the wall.

The momentum transfer rate per unit area gives the pressure: $\(P = \frac{1}{A} \sum_{\text{molecules}} 2p_x \cdot \frac{v_x}{2}\)$

For a gas in equilibrium with isotropic velocity distribution: $\(P = n\langle p_x v_x \rangle\)$

Since \(p_x = f(v^2)v_x\): $\(P = n\langle f(v^2)v_x \cdot v_x \rangle = n\langle v_x^2 f(v^2) \rangle\)$

By isotropy, \(\langle v_x^2 \rangle = \frac{1}{3}\langle v^2 \rangle\), and since \(f(v^2)\) depends only on the magnitude: $\(P = \frac{n}{3}\langle v^2 f(v^2) \rangle\)$

With \(\vec{p} = f(v^2)\vec{v}\), we have \(p = |\vec{p}| = f(v^2)v\), so: $\(v_x p_x = v_x \cdot \frac{p_x}{v} \cdot p = \frac{v_x^2}{v} \cdot p\)$

Therefore: $\(P = n\langle v_x p_x \rangle = n\left\langle \frac{v_x^2}{v} p \right\rangle\)$

By symmetry and using \(\langle v_x^2 \rangle = \frac{1}{3}\langle v^2 \rangle\): $\(P = \frac{n}{3}\left\langle \frac{v^2}{v} p \right\rangle = \frac{n}{3}\langle v p \rangle = n\langle v_x p_x \rangle\)$

Question 2

For a relativistic gas, the momentum-velocity relation is: $\(\vec{p} = \gamma m \vec{v} = \frac{m\vec{v}}{\sqrt{1-v^2/c^2}}\)$

Using rotational invariance, the pressure can be written as: $\(P = \frac{1}{3}n\langle \vec{v} \cdot \vec{p} \rangle\)$

Since \(\vec{v} \cdot \vec{p} = v \cdot p \cdot \cos(0) = vp\) (momentum parallel to velocity): $\(P = \frac{1}{3}n\langle vp \rangle\)$

For the relativistic case: $\(vp = v \cdot \gamma mv = \frac{mv^2}{\sqrt{1-v^2/c^2}}\)$

This can be rewritten using the relativistic energy-momentum relation: $\(E^2 = (pc)^2 + (mc^2)^2\)$

The kinetic energy is \(T = E - mc^2 = \gamma mc^2 - mc^2\), and: $\(vp = \frac{v}{c} \cdot pc = \frac{v}{c} \cdot \frac{E}{c}\sqrt{1-\frac{(mc^2)^2}{E^2}}\)$

After algebraic manipulation: $\(vp = \gamma mv^2 = (\gamma - 1)mc^2 + \frac{v^2}{c^2}\gamma mc^2\)$

Since \(\gamma - 1 = \frac{v^2/c^2}{1-v^2/c^2} \cdot \frac{1}{2} + O(v^4/c^4)\) and for an isotropic distribution: $\(\langle vp \rangle = \langle (\gamma - 1)mc^2 \rangle + \frac{1}{3}\langle p^2/(\gamma m) \rangle\)$

This simplifies to: $\(P = \frac{1}{3}n\langle vp \rangle = \frac{1}{3}n\langle \vec{v} \cdot \vec{p} \rangle\)$

Therefore, \(\alpha = \frac{1}{3}\).

Question 3

For ultrarelativistic particles, \(v \approx c\) and \(E = pc\).

From the general pressure formula: $\(P = \frac{1}{3}n\langle \vec{v} \cdot \vec{p} \rangle\)$

Since \(\vec{p} = (E/c)\hat{v}\) where \(\hat{v}\) is the unit vector in the direction of velocity: $\(\vec{v} \cdot \vec{p} = v \cdot \frac{E}{c} = \frac{vE}{c}\)$

For ultrarelativistic particles, \(v \approx c\): $\(\vec{v} \cdot \vec{p} \approx E\)$

Therefore: $\(P = \frac{1}{3}n\langle E \rangle\)$

Since the average energy per particle is \(\langle E \rangle\): $\(P = \frac{1}{3}n\langle E \rangle = \beta n\langle E \rangle\)$

Thus, \(\beta = \frac{1}{3}\).

Question 4

From the first law of thermodynamics for an adiabatic process: $\(dU = -PdV\)$

The total internal energy is \(U = N\langle E \rangle\), where \(N\) is the total number of particles.

Since \(n = N/V\): $\(U = nV\langle E \rangle\)$

From Question 3, \(P = \frac{1}{3}n\langle E \rangle\), so: $\(\langle E \rangle = \frac{3P}{n}\)$

Therefore: $\(U = nV \cdot \frac{3P}{n} = 3PV\)$

For an adiabatic process: $\(dU = d(3PV) = 3PdV + 3VdP = -PdV\)$

This gives: $\(3PdV + 3VdP = -PdV\)$ $\(4PdV + 3VdP = 0\)$ $\(\frac{dP}{P} + \frac{4}{3}\frac{dV}{V} = 0\)$

Integrating: $\(\ln P + \frac{4}{3}\ln V = \text{const}\)$ $\(PV^{4/3} = \text{const}\)$

Therefore, \(\gamma = \frac{4}{3}\) for ultrarelativistic gas.