Kinetic Theory — NCERT Solutions

Use the ideal gas equation with absolute pressure and absolute temperature.

Initial state

Gauge pressure $=15\,\text{atm}$, so absolute pressure:

$$P_1=15+1=16\,\text{atm}$$

Temperature:

$$T_1=27^\circ\text{C}=300\,\text{K}$$

Volume:

$$V=30\,\text{L}=30\times10^{-3}\,\text{m}^3$$

Number of moles initially:

$$\mu_1=\frac{P_1V}{RT_1}$$

Using $1\,\text{atm}=1.01\times10^5\,\text{Pa}$,

$$P_1=16\times1.01\times10^5\,\text{Pa}$$

$$\mu_1=\frac{16\times1.01\times10^5\times30\times10^{-3}}{8.31\times300}$$

$$\mu_1\approx19.5\,\text{mol}$$

Final state

Gauge pressure $=11\,\text{atm}$, so

$$P_2=12\,\text{atm}$$

Temperature:

$$T_2=17^\circ\text{C}=290\,\text{K}$$

So

$$\mu_2=\frac{P_2V}{RT_2}$$

$$\mu_2=\frac{12\times1.01\times10^5\times30\times10^{-3}}{8.31\times290}$$

$$\mu_2\approx15.4\,\text{mol}$$

Moles withdrawn:

$$\Delta\mu=19.5-15.4=4.1\,\text{mol}$$

Mass withdrawn:

$$m=\Delta\mu\times32\,\text{g mol}^{-1}$$

$$m\approx4.1\times32\approx131\,\text{g}$$

So the mass of oxygen taken out is about $1.3\times10^2$ g.

Use the ideal gas law, so for the same amount of gas:

$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$

For the bubble at the bottom:

• $V_1=1.0\,\text{cm}^3$
• depth $=40\,\text{m}$, so pressure there is approximately

$$P_1=1+\frac{40}{10}=5\,\text{atm}$$

• $T_1=12^\circ\text{C}=285\,\text{K}$

At the surface:

• $P_2=1\,\text{atm}$
• $T_2=35^\circ\text{C}=308\,\text{K}$

Therefore,

$$V_2=V_1\frac{P_1}{P_2}\frac{T_2}{T_1}$$

$$V_2=1.0\times5\times\frac{308}{285}$$

$$V_2\approx5.4\,\text{cm}^3$$

So the bubble grows to about $5.4\,\text{cm}^3$.

Use the ideal gas equation:

$$PV=\mu RT$$

Given:

• $P=1\,\text{atm}=1.01\times10^5\,\text{Pa}$
• $V=25.0\,\text{m}^3$
• $T=27^\circ\text{C}=300\,\text{K}$
• $R=8.31\,\text{J mol}^{-1}\text{K}^{-1}$

Number of moles:

$$\mu=\frac{PV}{RT}=\frac{1.01\times10^5\times25.0}{8.31\times300}$$

$$\mu\approx1.01\times10^3\,\text{mol}$$

Number of molecules:

$$N=\mu N_A=(1.01\times10^3)(6.02\times10^{23})$$

$$N\approx6.1\times10^{26}$$

So the room contains about $6.1\times10^{26}$ molecules.

For an ideal gas atom, the average thermal energy per molecule is

$$\langle E\rangle=\frac{3}{2}k_BT$$

with $k_B=1.38\times10^{-23}\,\text{J K}^{-1}$.

(i) Room temperature, $27^\circ\text{C}=300\,\text{K}$

$$\langle E\rangle=\frac{3}{2}(1.38\times10^{-23})(300)$$

$$\langle E\rangle=6.21\times10^{-21}\,\text{J}$$

(ii) Surface of the Sun, $6000\,\text{K}$

$$\langle E\rangle=\frac{3}{2}(1.38\times10^{-23})(6000)$$

$$\langle E\rangle=1.24\times10^{-19}\,\text{J}$$

(iii) $10^7\,\text{K}$

$$\langle E\rangle=\frac{3}{2}(1.38\times10^{-23})(10^7)$$

$$\langle E\rangle=2.07\times10^{-16}\,\text{J}$$

So the average thermal energies are $6.21\times10^{-21}$ J, $1.24\times10^{-19}$ J, and $2.07\times10^{-16}$ J.

At the same temperature and pressure, equal volumes of ideal gases contain equal numbers of molecules by Avogadro’s law. Since the vessels have equal capacity, they contain equal numbers of molecules.

But the rms speed is

$$v_{\mathrm{rms}}=\sqrt{\frac{3k_BT}{m}}$$

So at the same temperature, lighter molecules move faster.

• Neon is lightest among Ne, Cl$_2$, and UF$_6$
• therefore its rms speed is the largest

So:

• Yes, the numbers of molecules are equal.
• No, the rms speeds are not equal.
• The largest $v_{\mathrm{rms}}$ is for neon.

For equal rms speeds,

$$\sqrt{\frac{3k_BT_{Ar}}{m_{Ar}}}=\sqrt{\frac{3k_BT_{He}}{m_{He}}}$$

So

$$\frac{T_{Ar}}{m_{Ar}}=\frac{T_{He}}{m_{He}}$$

Thus,

$$T_{Ar}=T_{He}\frac{m_{Ar}}{m_{He}}$$

Given:

• $T_{He}=-20^\circ\text{C}=253\,\text{K}$
• $m_{Ar}=39.9u$
• $m_{He}=4.0u$

So

$$T_{Ar}=253\times\frac{39.9}{4.0}$$

$$T_{Ar}\approx253\times9.975\approx2524\,\text{K}$$

Convert to Celsius:

$$2524-273\approx2251^\circ\text{C}$$

So the argon cylinder must be at about $2.5\times10^3\,\text{K}$ or $2250^\circ\text{C}$.

For nitrogen, take molecular radius $r\approx1.0\,\AA$, so diameter

$$d=2r=2.0\times10^{-10}\,\text{m}$$

Given:

• pressure $P=2.0\,\text{atm}=2.02\times10^5\,\text{Pa}$
• temperature $T=17^\circ\text{C}=290\,\text{K}$
• molar mass of $N_2 = 28.0\,u$

1) Number density

Using

$$P=nk_BT$$

$$n=\frac{P}{k_BT}=\frac{2.02\times10^5}{(1.38\times10^{-23})(290)}$$

$$n\approx5.0\times10^{25}\,\text{m}^{-3}$$

2) Mean free path

Using

$$l=\frac{1}{\sqrt2\,n\pi d^2}$$

$$d^2=(2.0\times10^{-10})^2=4.0\times10^{-20}$$

$$l=\frac{1}{1.414\times(5.0\times10^{25})\times\pi\times4.0\times10^{-20}}$$

$$l\approx1.1\times10^{-7}\,\text{m}$$

So the mean free path is about $1\times10^{-7}$ m.

3) Collision frequency

First find rms speed:

$$v_{\mathrm{rms}}=\sqrt{\frac{3k_BT}{m}}$$

Mass of one $N_2$ molecule:

$$m=\frac{28.0\times10^{-3}}{6.02\times10^{23}}\approx4.65\times10^{-26}\,\text{kg}$$

Then

$$v_{\mathrm{rms}}=\sqrt{\frac{3(1.38\times10^{-23})(290)}{4.65\times10^{-26}}}$$

$$v_{\mathrm{rms}}\approx5.2\times10^2\,\text{m s}^{-1}$$

Collision time:

$$\tau=\frac{l}{v_{\mathrm{rms}}}\approx\frac{1.1\times10^{-7}}{5.2\times10^2}$$

$$\tau\approx2.1\times10^{-10}\,\text{s}$$

Collision frequency:

$$\nu=\frac{1}{\tau}\approx4.8\times10^9\,\text{s}^{-1}$$

Comparison

• Time between collisions $\approx 2\times10^{-10}$ s
• Collision frequency $\approx 5\times10^9\,\text{s}^{-1}$

So the molecule moves freely for about $10^{-10}$ s between collisions, covering a distance of about $10^{-7}$ m.