Kinetic TheoryClass 11 Physics NCERT Solutions

Given: Diameter of an oxygen molecule, $d = 3 \AA = 3 \times 10^{-10} \text{ m}$. Conditions are Standard Temperature and Pressure (STP).

To Find: The fraction of molecular volume to the actual volume occupied by oxygen gas.

Formula: Volume of one molecule, $V_{\text{molecule}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (\frac{d}{2})^3 = \frac{1}{6}\pi d^3$. At STP, 1 mole of any ideal gas occupies a volume of 22.4 litres. This is the actual volume, $V_{\text{actual}}$. Number of molecules in 1 mole = Avogadro's number, $N_A = 6.023 \times 10^{23}$. Total molecular volume, $V_{\text{molecular}} = N_A \times V_{\text{molecule}}$.

Calculation: Radius of an oxygen molecule, $r = \frac{d}{2} = \frac{3 \times 10^{-10}}{2} = 1.5 \times 10^{-10} \text{ m}$. Volume of one oxygen molecule: $V_{\text{molecule}} = \frac{4}{3}\pi r^3 = \frac{4}{3} \times 3.14 \times (1.5 \times 10^{-10})^3 \text{ m}^3$ $V_{\text{molecule}} = 1.413 \times 10^{-29} \text{ m}^3$

Total volume of molecules in 1 mole of oxygen gas (molecular volume): $V_{\text{molecular}} = N_A \times V_{\text{molecule}} = (6.023 \times 10^{23}) \times (1.413 \times 10^{-29}) \text{ m}^3$ $V_{\text{molecular}} = 8.51 \times 10^{-6} \text{ m}^3$

Actual volume occupied by 1 mole of oxygen gas at STP: $V_{\text{actual}} = 22.4 \text{ litres} = 22.4 \times 10^{-3} \text{ m}^3$

Fraction of molecular volume to actual volume: $\text{Fraction} = \frac{V_{\text{molecular}}}{V_{\text{actual}}} = \frac{8.51 \times 10^{-6} \text{ m}^3}{22.4 \times 10^{-3} \text{ m}^3}$ $\text{Fraction} = 0.3799 \times 10^{-3} \approx 3.8 \times 10^{-4}$

Final Answer: The fraction of the molecular volume to the actual volume occupied by oxygen gas at STP is approximately $3.8 \times 10^{-4}$.

Given: Conditions are Standard Temperature and Pressure (STP). Pressure, $P = 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}$. Temperature, $T = 0^{\circ} \mathrm{C} = 273.15 \text{ K}$. Number of moles, $\mu = 1 \text{ mol}$. Universal gas constant, $R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}$.

To Find: The molar volume, $V$.

Formula: The ideal gas equation is: $PV = \mu RT$

Calculation: Rearranging the formula to find the volume $V$: $V = \frac{\mu RT}{P}$ Substituting the given values: $V = \frac{(1 \text{ mol}) \times (8.314 \text{ J mol}^{-1} \text{K}^{-1}) \times (273.15 \text{ K})}{1.013 \times 10^5 \text{ Pa}}$ $V = \frac{2271.06}{1.013 \times 10^5} \text{ m}^3$ $V = 22.419 \times 10^{-3} \text{ m}^3$ Since $1 \text{ m}^3 = 1000$ litres, we convert the volume to litres: $V = (22.419 \times 10^{-3}) \times 1000 \text{ litres}$ $V \approx 22.4 \text{ litres}$

Final Answer: Thus, it is shown that the molar volume of an ideal gas at STP is 22.4 litres.

(a) For an ideal gas, the ideal gas equation is $PV = \mu RT$. This can be written as $\frac{PV}{T} = \mu R$. Since the number of moles $\mu$ and the universal gas constant $R$ are constant for a given sample of gas, the term $\frac{PV}{T}$ is a constant and does not depend on pressure $P$. The dotted plot is a straight line parallel to the pressure axis, which signifies the behavior of an ideal gas.

(b) Real gases deviate from ideal gas behavior. However, they approach ideal gas behavior at high temperatures and low pressures. In the given plot, the curve at temperature $T_1$ is closer to the ideal gas line (the dotted plot) than the curve at temperature $T_2$. This indicates that the gas behaves more like an ideal gas at temperature $T_1$. Therefore, it is true that $T_1 > T_2$.

(c) The curves meet on the y-axis where the pressure $P$ approaches zero. In the limit of very low pressure, a real gas behaves like an ideal gas. Therefore, the value of $\frac{PV}{T}$ at this point is equal to $\mu R$. Given: Mass of oxygen gas, $m = 1.00 \times 10^{-3} \text{ kg}$. Molecular mass of oxygen, $M_{O_2} = 32.0 \text{ u} = 32.0 \times 10^{-3} \text{ kg/mol}$. Gas constant, $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$.

Calculation: Number of moles of oxygen, $\mu_{O_2} = \frac{m}{M_{O_2}} = \frac{1.00 \times 10^{-3} \text{ kg}}{32.0 \times 10^{-3} \text{ kg/mol}} = 0.03125 \text{ mol}$. Value of $\frac{PV}{T} = \mu R = (0.03125 \text{ mol}) \times (8.31 \text{ J mol}^{-1} \text{K}^{-1}) = 0.2597 \text{ J K}^{-1}$. So, the value of $PV/T$ where the curves meet the y-axis is $0.26 \text{ J K}^{-1}$.

(d) No, we would not get the same value of $PV/T$. The value of $PV/T$ depends on the number of moles ($\mu$). Given: Mass of hydrogen, $m_{H_2} = 1.00 \times 10^{-3} \text{ kg}$. Molecular mass of hydrogen, $M_{H_2} = 2.02 \text{ u} = 2.02 \times 10^{-3} \text{ kg/mol}$.

Calculation for Hydrogen: Number of moles of hydrogen, $\mu_{H_2} = \frac{1.00 \times 10^{-3} \text{ kg}}{2.02 \times 10^{-3} \text{ kg/mol}} = 0.495 \text{ mol}$. The value of $PV/T$ for hydrogen would be $\mu_{H_2} R = 0.495 \times 8.31 = 4.11 \text{ J K}^{-1}$, which is different.

To get the same value of $PV/T$, the number of moles of hydrogen must be the same as the number of moles of oxygen. $\mu_{H_2} = \mu_{O_2} = 0.03125 \text{ mol}$ Mass of hydrogen required, $m'_{H_2} = \mu_{H_2} \times M_{H_2}$. $m'_{H_2} = 0.03125 \text{ mol} \times 2.02 \times 10^{-3} \text{ kg/mol} = 6.3125 \times 10^{-5} \text{ kg}$

Final Answer: The mass of hydrogen that yields the same value of $PV/T$ is $6.31 \times 10^{-5} \text{ kg}$.

Given: Volume of the cylinder, $V = 30 \text{ litres} = 30 \times 10^{-3} \text{ m}^3$. Initial gauge pressure, $P_{g1} = 15 \text{ atm}$. Initial temperature, $T_1 = 27^{\circ} \mathrm{C} = 27 + 273 = 300 \text{ K}$. Final gauge pressure, $P_{g2} = 11 \text{ atm}$. Final temperature, $T_2 = 17^{\circ} \mathrm{C} = 17 + 273 = 290 \text{ K}$. Universal gas constant, $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$. Molecular mass of oxygen, $M_{O_2} = 32 \text{ u} = 32 \times 10^{-3} \text{ kg/mol}$. Atmospheric pressure, $P_a = 1 \text{ atm}$.

To Find: The mass of oxygen taken out of the cylinder, $\Delta m = m_1 - m_2$.

Formula: The ideal gas equation is $PV = \mu RT = \frac{m}{M}RT$. Therefore, mass $m = \frac{PVM}{RT}$. Absolute Pressure = Gauge Pressure + Atmospheric Pressure.

Calculation: Convert pressures from atm to Pa: $1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}$.

Initial State: Initial absolute pressure, $P_1 = P_{g1} + P_a = 15 + 1 = 16 \text{ atm} = 16 \times 1.013 \times 10^5 \text{ Pa} = 1.6208 \times 10^6 \text{ Pa}$. Initial mass of oxygen, $m_1$: $m_1 = \frac{P_1 V M}{R T_1} = \frac{(1.6208 \times 10^6 \text{ Pa}) \times (30 \times 10^{-3} \text{ m}^3) \times (32 \times 10^{-3} \text{ kg/mol})}{(8.31 \text{ J mol}^{-1} \text{K}^{-1}) \times (300 \text{ K})}$ $m_1 = \frac{1555.968}{2493} = 0.6241 \text{ kg}$

Final State: Final absolute pressure, $P_2 = P_{g2} + P_a = 11 + 1 = 12 \text{ atm} = 12 \times 1.013 \times 10^5 \text{ Pa} = 1.2156 \times 10^6 \text{ Pa}$. Final mass of oxygen, $m_2$: $m_2 = \frac{P_2 V M}{R T_2} = \frac{(1.2156 \times 10^6 \text{ Pa}) \times (30 \times 10^{-3} \text{ m}^3) \times (32 \times 10^{-3} \text{ kg/mol})}{(8.31 \text{ J mol}^{-1} \text{K}^{-1}) \times (290 \text{ K})}$ $m_2 = \frac{1167.0}{2409.9} = 0.4842 \text{ kg}$

Mass of oxygen taken out: $\Delta m = m_1 - m_2 = 0.6241 \text{ kg} - 0.4842 \text{ kg} = 0.1399 \text{ kg}$

Final Answer: The estimated mass of oxygen taken out of the cylinder is $0.140 \text{ kg}$.

Given: At the bottom of the lake (State 1): Initial volume, $V_1 = 1.0 \text{ cm}^3 = 1.0 \times 10^{-6} \text{ m}^3$. Depth of the lake, $h = 40 \text{ m}$. Initial temperature, $T_1 = 12^{\circ} \mathrm{C} = 12 + 273 = 285 \text{ K}$.

At the surface of the lake (State 2): Final temperature, $T_2 = 35^{\circ} \mathrm{C} = 35 + 273 = 308 \text{ K}$.

Other constants: Atmospheric pressure, $P_a = 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}$. Density of water, $\rho = 1000 \text{ kg/m}^3$. Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$.

To Find: The final volume of the bubble, $V_2$.

Formula: The combined gas law for a fixed amount of gas is: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ Pressure at depth $h$ is $P_1 = P_a + h\rho g$. Pressure at the surface is $P_2 = P_a$.

Calculation: Pressure at the bottom ($P_1$): Pressure due to water column = $h\rho g = (40 \text{ m}) \times (1000 \text{ kg/m}^3) \times (9.8 \text{ m/s}^2) = 3.92 \times 10^5 \text{ Pa}$. $P_1 = P_a + h\rho g = (1.013 \times 10^5 \text{ Pa}) + (3.92 \times 10^5 \text{ Pa}) = 4.933 \times 10^5 \text{ Pa}$

Pressure at the surface ($P_2$): $P_2 = P_a = 1.013 \times 10^5 \text{ Pa}$

Using the combined gas law to find $V_2$: $V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}$ $V_2 = (1.0 \times 10^{-6} \text{ m}^3) \times \frac{4.933 \times 10^5 \text{ Pa}}{1.013 \times 10^5 \text{ Pa}} \times \frac{308 \text{ K}}{285 \text{ K}}$ $V_2 = (1.0 \times 10^{-6}) \times (4.87) \times (1.08) \text{ m}^3$ $V_2 = 5.26 \times 10^{-6} \text{ m}^3$ Converting back to $\text{cm}^3$: $V_2 = 5.26 \text{ cm}^3$

Final Answer: The air bubble grows to a volume of $5.26 \text{ cm}^3$ when it reaches the surface.

Given: Volume of the room, $V = 25.0 \text{ m}^3$. Temperature, $T = 27^{\circ} \mathrm{C} = 27 + 273 = 300 \text{ K}$. Pressure, $P = 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}$. Boltzmann constant, $k_B = 1.38 \times 10^{-23} \text{ J K}^{-1}$.

To Find: The total number of air molecules, $N$.

Formula: The ideal gas equation in terms of the number of molecules is: $PV = N k_B T$

Calculation: Rearranging the formula to solve for $N$: $N = \frac{PV}{k_B T}$ Substituting the given values: $N = \frac{(1.013 \times 10^5 \text{ Pa}) \times (25.0 \text{ m}^3)}{(1.38 \times 10^{-23} \text{ J K}^{-1}) \times (300 \text{ K})}$ $N = \frac{2.5325 \times 10^6}{4.14 \times 10^{-21}}$ $N = 0.6117 \times 10^{27} = 6.117 \times 10^{26}$

Final Answer: The total number of air molecules in the room is approximately $6.12 \times 10^{26}$.

Part 1: Number of Molecules Yes, the vessels contain an equal number of respective molecules. According to Avogadro's law, equal volumes of all gases at the same temperature and pressure contain the same number of molecules. This can also be seen from the ideal gas equation, $PV = N k_B T$. Since the pressure ($P$), volume ($V$), and temperature ($T$) are the same for all three vessels, the number of molecules ($N$) must also be the same.

Part 2: Root Mean Square (rms) Speed No, the root mean square speed of the molecules is not the same in the three cases. The formula for the rms speed is: $v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}$ where $m$ is the mass of a single molecule. Although the temperature ($T$) is the same for all three gases, the molecular masses ($m$) are different:

Part 3: Largest rms Speed From the formula $v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}$, we can see that the rms speed is inversely proportional to the square root of the molecular mass ($v_{\text{rms}} \propto \frac{1}{\sqrt{m}}$). This means the gas with the lowest molecular mass will have the highest rms speed. Comparing the molecular masses: m_{\text{Ne}} < m_{\text{Cl}_2} < m_{\text{UF}_6}} Therefore, neon (Ne) will have the largest root mean square speed.

Given: Atomic mass of Argon, $M_{Ar} = 39.9 \text{ u}$. Atomic mass of Helium, $M_{He} = 4.0 \text{ u}$. Temperature of Helium, $T_{He} = -20^{\circ} \mathrm{C} = -20 + 273.15 = 253.15 \text{ K}$. Condition: $(v_{\text{rms}})_{Ar} = (v_{\text{rms}})_{He}$.

To Find: The temperature of Argon, $T_{Ar}$.

Formula: The root mean square (rms) speed of a gas molecule is given by: $v_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_B T}{m}}$ where $M$ is the molar mass and $m$ is the mass of one atom.

Calculation: According to the given condition: $\sqrt{\frac{3RT_{Ar}}{M_{Ar}}} = \sqrt{\frac{3RT_{He}}{M_{He}}}$ Squaring both sides and cancelling the common term $3R$: $\frac{T_{Ar}}{M_{Ar}} = \frac{T_{He}}{M_{He}}$ Rearranging to solve for $T_{Ar}$: $T_{Ar} = T_{He} \times \frac{M_{Ar}}{M_{He}}$ Substituting the values: $T_{Ar} = (253.15 \text{ K}) \times \frac{39.9 \text{ u}}{4.0 \text{ u}}$ $T_{Ar} = 253.15 \times 9.975 \text{ K}$ $T_{Ar} = 2525.16 \text{ K}$

To express the temperature in degrees Celsius: $T_{Ar} (^{\circ} \mathrm{C}) = 2525.16 - 273.15 = 2252.01^{\circ} \mathrm{C}$

Final Answer: The temperature of the argon gas must be $2525.16 \text{ K}$ or $2252.01^{\circ} \mathrm{C}$.

Given: Pressure, $P = 2.0 \text{ atm} = 2.0 \times 1.013 \times 10^5 \text{ Pa} = 2.026 \times 10^5 \text{ Pa}$. Temperature, $T = 17^{\circ} \mathrm{C} = 17 + 273 = 290 \text{ K}$. Radius of a nitrogen molecule, $r = 1.0 \AA = 1.0 \times 10^{-10} \text{ m}$. Diameter of a nitrogen molecule, $d = 2r = 2.0 \times 10^{-10} \text{ m}$. Molecular mass of nitrogen, $M_{N_2} = 28.0 \text{ u} = 28.0 \times 10^{-3} \text{ kg/mol}$. Boltzmann constant, $k_B = 1.38 \times 10^{-23} \text{ J K}^{-1}$. Avogadro's number, $N_A = 6.023 \times 10^{23} \text{ mol}^{-1}$.

Formulas: Number density, $n = \frac{P}{k_B T}$. Mean free path, $l = \frac{1}{\sqrt{2} n \pi d^2}$. Mass of one molecule, $m = \frac{M_{N_2}}{N_A}$. RMS speed, $v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}$. Collision frequency, $f = \frac{v_{\text{rms}}}{l}$. Time between collisions (free time), $\tau = \frac{1}{f} = \frac{l}{v_{\text{rms}}}$. Collision time, $t_c = \frac{d}{v_{\text{rms}}}$.

Calculation: 1. Mean Free Path ($l$) First, calculate the number density $n$: $n = \frac{2.026 \times 10^5 \text{ Pa}}{(1.38 \times 10^{-23} \text{ J K}^{-1}) \times (290 \text{ K})} = \frac{2.026 \times 10^5}{4.002 \times 10^{-21}} = 5.06 \times 10^{25} \text{ m}^{-3}$ Now, calculate the mean free path $l$: $l = \frac{1}{\sqrt{2} \times (5.06 \times 10^{25} \text{ m}^{-3}) \times \pi \times (2.0 \times 10^{-10} \text{ m})^2}$ $l = \frac{1}{1.414 \times (5.06 \times 10^{25}) \times 3.14 \times (4 \times 10^{-20})} = \frac{1}{8.99 \times 10^6} = 1.11 \times 10^{-7} \text{ m}$

2. Collision Frequency ($f$) First, calculate the mass of one nitrogen molecule $m$: $m = \frac{28.0 \times 10^{-3} \text{ kg/mol}}{6.023 \times 10^{23} \text{ mol}^{-1}} = 4.65 \times 10^{-26} \text{ kg}$ Next, calculate the rms speed $v_{\text{rms}}$: $v_{\text{rms}} = \sqrt{\frac{3 \times (1.38 \times 10^{-23} \text{ J K}^{-1}) \times (290 \text{ K})}{4.65 \times 10^{-26} \text{ kg}}} = \sqrt{\frac{1.2006 \times 10^{-20}}{4.65 \times 10^{-26}}} = \sqrt{2.58 \times 10^5} = 508 \text{ m/s}$ Finally, calculate the collision frequency $f$: $f = \frac{v_{\text{rms}}}{l} = \frac{508 \text{ m/s}}{1.11 \times 10^{-7} \text{ m}} = 4.58 \times 10^9 \text{ s}^{-1}$

3. Comparison of Times Time between collisions (free time): $\tau = \frac{l}{v_{\text{rms}}} = \frac{1.11 \times 10^{-7} \text{ m}}{508 \text{ m/s}} = 2.18 \times 10^{-10} \text{ s}$ Collision time (time taken to travel its own diameter): $t_c = \frac{d}{v_{\text{rms}}} = \frac{2.0 \times 10^{-10} \text{ m}}{508 \text{ m/s}} = 3.94 \times 10^{-13} \text{ s}$ Ratio of free time to collision time: $\frac{\tau}{t_c} = \frac{2.18 \times 10^{-10} \text{ s}}{3.94 \times 10^{-13} \text{ s}} \approx 553$