CBSE Class 11 Physics Kinetic Theory of Gases Theory and Examples

1. ASSUMPTION OF KINETIC THEORY

CBSE Class 12 Physics: Kinetic Theory of Gases, Theory and Examples

1. Assumptions of kinetic theory

(i) Assumptions regarding the molecule

1. Every gas consists of extremely small particles known as molecules. The molecules of a given gas are all identical but are different from those of another gas.
2. The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses.
3. Their size is negligible in comparison to the intermolecular distance (\(10^{-9}\) m).

(ii) Assumptions regarding volume

The volume of molecules is negligible in comparison to the volume of the gas. (The volume of the molecules is only 0.014 % of the volume of the gas.)

(iii) Assumptions regarding motion

1. Molecules of a gas keep moving randomly in all possible directions with all possible velocities.
2. The speed of gas molecules lies between zero and infinity (very high speed).
3. The number of molecules moving with the most probable speed is maximum.

(iv) Assumptions regarding collision

1. The gas molecules keep on colliding among themselves as well as with the walls of the containing vessel. These collisions are perfectly elastic (i.e. the total energy before collision = total energy after the collision).
2. Molecules move in a straight line with constant speed during successive collisions.
3. The distance covered by the molecules between two successive collisions is known as the free path, and the mean of all free paths is known as the mean free path.
4. The time spent in a collision between two molecules is negligible in comparison to the time between two successive collisions.
5. The number of collisions per unit volume in a gas remains constant.

(v) Assumptions regarding force

1. No attractive or repulsive force acts between gas molecules.
2. Gravitational attraction among the molecules is ineffective due to extremely small masses and very high speed of the molecules.

(vi) Assumptions regarding pressure

Molecules constantly collide with the walls of the container, due to which their momentum changes. This change in momentum is transferred to the walls of the container. Consequently, pressure is exerted by gas molecules on the walls of the container.

(vii) Assumptions regarding density

The density of a gas is constant at all points of the container.

2. Some definitions

(i) Gram mol or kilogram mol

1. The quantity of matter in which the number of molecules is equal to Avogadro's number is defined as a gram mol.
2. The molecular weight of any substance expressed in kilograms is defined as 1 kilogram mol of that substance.
3. The molecular weight of any substance expressed in grams is defined as 1 gram mol of that substance.
4. 1 kg mol of H₂ = 2 kg, 1 kg mol of N₂ = 28 kg, 1 kg mol of O₂ = 32 kg, 1 kg mol of CO₂ = 44 kg.
5. Number of molecules in 1 mol \(=\dfrac{\text{total number of molecules}}{\text{number of moles}}\)
6. Number of moles \(=\dfrac{\text{mass of gas}}{\text{molecular weight of gas}}\), or \(n=\dfrac{m}{M}\)
7. The mass of 1 mol of a gas is equal to its molecular weight.

(ii) Avogadro's number (N_A)

1. The number of molecules present in 1 mol of a gas is defined as Avogadro's number.
2. \(N_A=6.02\times10^{23}\) per gram mol \(=6.02\times10^{26}\) per kg mol.

(iii) Molar volume and molar mass (V_m)

1. Molar volume \(V_m\): the volume of 1 mol of a gas is known as molar volume.
2. \(V_m=\dfrac{V}{n}\)
3. The unit of \(V_m\) is m³/mol.

Molecular weight (M)

1. The mass of 1 mol of matter is its molecular weight.
2. \(M=mN_A\) (\(m\) = mass of a single molecule)

(iv) Meaning of NTP: NTP means normal temperature and pressure.

1. Temperature at NTP = 0 °C = 273 K
2. Pressure at NTP = 76 cm of Hg column \(=1.013\times10^{5}\) N/m² (Pa) = 1 atmosphere.
3. Volume of 1 mol of gas at NTP = 22.4 litre.

(v) Absolute zero temperature (0 K)

1. The minimum possible temperature at which all the gas molecules come to rest is defined as absolute zero temperature (i.e. 0 K).
2. The temperature at which the kinetic energy of gas molecules becomes zero is defined as absolute zero temperature.
3. At this temperature the volume of the gas becomes zero.

Symbols used in this chapter

• \(R\) = universal gas constant = 8.314 J/mol-K = 0.082 litre-atm/mol-K = 1.986 cal/mol-K
• \(M\) = molecular weight in kg = mass of \(6.023\times10^{23}\) molecules of the gas
• \(m\) = mass of each molecule in kg \(=M/N_0\)
• \(N_0\) = Avogadro's constant \(=6.023\times10^{23}\)
• \(k\) = Boltzmann's constant \(=R/N_0=1.38\times10^{-23}\) J/K
• \(P\) = pressure of the gas in N/m²
• \(T\) = temperature in kelvin
• \(N\) = total number of molecules
• \(d\) = density of the gas \(=nM/V\)
• \(n\) = number of moles

3. Different velocities of molecules of a gas

1. Instantaneous velocity \((\vec v)\): velocity of a molecule at any instant of time.
2. Instantaneous speed \((|\vec v|)\): speed of a molecule at any instant of time.
3. Average velocity \((\bar v)\):

\[\langle\vec v\rangle=\bar{\vec v}=\frac{\sum\vec v}{N}=\frac{\vec v_1+\vec v_2+\dots+\vec v_N}{N}=0\]

This is zero because the molecules are in random motion.

Average speed \((|\bar v|)\):

\[\langle|\vec v|\rangle=\frac{\sum|\vec v|}{N}=\frac{|\vec v_1|+|\vec v_2|+\dots+|\vec v_N|}{N}=\sqrt{\frac{8RT}{\pi M}}=1.59\sqrt{\frac{RT}{M}}=1.59\sqrt{\frac{kT}{m}}\]

Most probable speed \((v_{mp})\): the speed with which the maximum number of molecules travel.

\[v_{mp}=\sqrt{\frac{2RT}{M}}=\sqrt{\frac{2kT}{m}}\]

Mean square velocity \((\overline{v^2})\): the average of the squares of the instantaneous velocities of all the molecules.

\[\langle v^2\rangle=\overline{v^2}=\frac{\sum v^2}{N}=\frac{v_1^2+v_2^2+\dots+v_N^2}{N}\]

Root mean square speed (RMS speed) \((v_{rms})\):

\[v_{rms}=\sqrt{\frac{\sum v^2}{N}}=\sqrt{\frac{v_1^2+v_2^2+\dots+v_N^2}{N}}\]

\[v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3PV}{nM}}=\sqrt{\frac{3P}{d}}\]

Velocity of sound in the gas:

\[v_s=\sqrt{\frac{\gamma RT}{M}},\qquad\text{where }\gamma=\frac{C_p}{C_v}\]

Examples based on different velocities of gas molecules

What is the rms speed of helium gas molecules at STP?

Answer: \(v_{rms}=\sqrt{\dfrac{3RT}{M}}\). At STP, \(T=273\) K, \(M=4\) g/mol \(=0.004\) kg/mol, \(R=8.31\) J/mol-K. \[v_{rms}=\sqrt{\frac{3\times8.31\times273}{0.004}}=\sqrt{1.7\times10^{6}}\approx1.3\times10^{3}\text{ m/s}\]

The rms speed of oxygen at room temperature is about 500 m/s. The rms speed of hydrogen at the same temperature is about

1. 125 m/s
2. 2000 m/s
3. 8000 m/s
4. 31 m/s

Answer: (b) 2000 m/s. At constant temperature \(v_{rms}\propto\dfrac1{\sqrt M}\), so \[\frac{(v_{rms})_{O}}{(v_{rms})_{H}}=\sqrt{\frac{M_H}{M_O}}=\sqrt{\frac{2}{32}}=\frac14\] \[(v_{rms})_H=4\times500=2000\text{ m/s}\]

The velocities of four molecules of a gas are \(\sqrt3\), 3, 4 and 6 m/s. The root mean square velocity will be

1. 4
2. 16
3. 8
4. \(13\sqrt3/4\)

Answer: (a) 4 m/s. \[v_{rms}=\left(\frac{v_1^2+v_2^2+v_3^2+v_4^2}{4}\right)^{1/2}=\left(\frac{3+9+16+36}{4}\right)^{1/2}=\left(\frac{64}{4}\right)^{1/2}=4\text{ m/s}\]

The temperature of an ideal gas is increased from 120 K to 480 K. If the root mean square velocity of the gas molecules is \(v\) at 120 K, at 480 K it becomes

1. \(4v\)
2. \(2v\)
3. \(v/2\)
4. \(v/4\)

Answer: (b) \(2v\). Since \(v_{rms}\propto\sqrt T\), \[\frac{(v_{rms})_{T_2}}{(v_{rms})_{T_1}}=\sqrt{\frac{T_2}{T_1}}\Rightarrow(v_{rms})_{480}=v\sqrt{\frac{480}{120}}=2v\]

The root mean square (rms) speed of oxygen molecules (O₂) at a certain absolute temperature \(T\) is \(v\). If the temperature is doubled and oxygen gas dissociates into atomic oxygen, does the rms speed remain unchanged?

1. Yes
2. No
3. Nothing can be said
4. None of these

Answer: (b) No. \(v=\sqrt{3RT/M}\). When the temperature is doubled the new temperature is \(T'=2T\), and when O₂ dissociates into atomic oxygen the molecular weight becomes \(M'=M/2\). The new rms speed is \[v'=\sqrt{\frac{3RT'}{M'}}=\sqrt{\frac{3R\cdot2T}{M/2}}=2\sqrt{\frac{3RT}{M}}\] \[\therefore\;v'=2v\]

4. Maxwell's law of molecular velocity distribution

1. Molecules can have all possible velocities in the range from 0 to \(\infty\).
2. The curve between the number of molecules per unit velocity range and the velocity at different temperatures is as shown.

1. The number of molecules belonging to a finite velocity range is constant, however the velocity of each molecule is changing.

Results

1. Different molecules have different velocities.
2. As the temperature increases, the number of molecules having higher velocities increases.
3. At a given temperature, the velocity which the maximum number of molecules possess is called the most probable velocity (\(v_{mp}\)).
4. With the increase in temperature, the most probable velocity increases.
5. The area under the curve represents the total number of molecules.

5. Calculation of pressure of a gas

1. Consider an ideal gas enclosed in a cubical vessel of edge \(L\).

2. Consider a molecule moving with velocity \(\vec v=v_x\hat i+v_y\hat j+v_z\hat k\) in any direction.
3. This molecule collides with the shaded wall with velocity \(v_x\). Before the collision its momentum is \(mv_x\) and after the collision it is \(-mv_x\) (assuming an elastic collision). So the change in momentum of the molecule \(=-mv_x-mv_x=-2mv_x\).
4. By the law of conservation of momentum, the change in momentum of the wall is equal in magnitude but opposite in direction to that of the molecule: \(\Delta P=2mv_x\).
5. The distance travelled parallel to the x-direction between two successive collisions with the shaded wall is \(2L\). Thus the time between two successive collisions is \(\Delta t=\dfrac{2L}{v_x}\), and the number of collisions per second is \(n=\dfrac{v_x}{2L}\).
6. The momentum imparted per unit time to the wall by this molecule is \[\Delta F=n\Delta P=\frac{v_x}{2L}\cdot2mv_x=\frac{mv_x^2}{L}\]
7. The total force on the wall is \[F=\sum\frac{mv_x^2}{L}=\frac mL\sum v_x^2\]
8. Assuming the average velocity in all directions to be equal, \[\sum v_x^2=\sum v_y^2=\sum v_z^2=\frac13\sum\left(v_x^2+v_y^2+v_z^2\right)=\frac13\sum v^2\] so \(F=\dfrac13\dfrac mL\sum v^2\).
9. If \(N\) is the total number of molecules in the sample, \[F=\frac13\frac mL\,N\,\frac{\sum v^2}{N}\] Pressure on the wall = force/area: \[P=\frac F{L^2}=\frac13\frac{mN}{L^3}\frac{\sum v^2}{N}=\frac13\frac{m_0}{V}\frac{\sum v^2}{N}\] where \(m_0=mN\) is the total mass of the gas, \(V=L^3\) is the volume of the vessel and \(m\) is the mass of each molecule. Hence \[P=\frac13\,d\left(\frac{\sum v^2}{N}\right)\quad\left(d=\text{density}=\frac{m_0}{V}\right)\] \[\Rightarrow\;\mathbf{P=\tfrac13\,d\,v_{rms}^2}\quad\left(\because v_{rms}=\sqrt{\tfrac{\sum v^2}{N}}\right)\] or \(PV=\dfrac13m_0v_{rms}^2\) or \(PV=\dfrac13Nm\,v_{rms}^2\).
10. The energy of each molecule \(=\dfrac12mv_{rms}^2\), so the total energy \(E_0=\dfrac12Nm\,v_{rms}^2\Rightarrow Nm\,v_{rms}^2=2E_0\). Hence \[PV=\frac23E_0\quad\text{or}\quad P=\frac23E\] where \(E\) = energy per unit volume.
11. Dependence of pressure: (a) \(P\propto n\) (total number of moles); (b) \(P\propto m\) (total mass of the gas); (c) \(P\propto v_{rms}^2\); (d) \(P\propto1/V\) (\(V\) = volume); (e) \(P\propto d\) (\(d\) = density); (f) \(P\propto E\) (\(E\) = energy per unit volume).

Examples based on calculation of pressure of a gas

A flask of \(10^{-3}\) m³ volume contains \(3\times10^{22}\) molecules of oxygen at a certain temperature. The mass of one molecule of oxygen is \(5.3\times10^{-26}\) kg and the rms velocity of its molecules at the same temperature is 400 m/s. Calculate the pressure of the gas.

Answer: \(P=\dfrac13\dfrac{mN}{V}v_{rms}^2\). Here \(V=10^{-3}\) m³, \(N=3\times10^{22}\), \(m=5.3\times10^{-26}\) kg, \(v_{rms}=400\) m/s. \[P=\frac13\times\frac{5.3\times10^{-26}\times3\times10^{22}\times(400)^2}{10^{-3}}=8.48\times10^{4}\text{ N/m}^2\]

The pressure of an ideal gas is written as \(P=\dfrac{2E}{3V}\). Here \(E\) refers to

1. translational kinetic energy
2. rotational kinetic energy
3. vibrational kinetic energy
4. total kinetic energy

Answer: (a) translational kinetic energy. The pressure is \(P=\dfrac13\dfrac{mN}{V}v_{rms}^2\), so \[E=\frac12mN\,v_{rms}^2=\frac12m\left(v_1^2+v_2^2+\dots+v_N^2\right)=\frac12mv_1^2+\frac12mv_2^2+\dots+\frac12mv_N^2\] This is the sum of the energies of all the molecules, which represents the translational kinetic energy.

Which of the following parameters is the same for molecules of all gases at a given temperature?

1. Mass
2. Speed
3. Momentum
4. Kinetic energy

Answer: (d) Kinetic energy. The average translational kinetic energy of a molecule is \(\dfrac32kT\) (\(\dfrac32RT\) per mole). \(R\) and \(k\) are universal constants and the temperature is the same for all gases, so the kinetic energy is the same for molecules of all gases at a given temperature.

A gas is filled in a container at pressure \(P_0\). If the mass of each molecule is halved and the speeds are doubled, the resulting pressure will be

1. \(4P_0\)
2. \(2P_0\)
3. \(P_0\)
4. \(4P_0/\sqrt2\)

Answer: (b) \(2P_0\). \(P_0=\dfrac13\dfrac{mN}{V}v_{rms}^2\) (\(m\) = mass of each molecule, \(V\) = volume of the gas). Now \(m'=m/2\) and \(v_{rms}'=2v_{rms}\), so \[P'=\frac13\frac{N(m/2)\,4v_{rms}^2}{V}=2\cdot\frac13\frac{Nm}{V}v_{rms}^2=2P_0\]

The ratio of the number of collisions per second at the walls of containers by H₂ and Ne gas molecules kept at the same volume and temperature is

1. 10 : 1
2. 1 : 10
3. \(1:\sqrt{10}\)
4. \(\sqrt{10}:1\)

Answer: (d) \(\sqrt{10}:1\). For the wall perpendicular to the x-axis the number of collisions per second is \(\dfrac{v_x}{2L}\), with \(v_x=v_{rms}/\sqrt3\). \(L\) and \(T\) are the same for both gases, so the ratio of the number of collisions equals the ratio of the speeds: \[\frac{v_{rms,H_2}}{v_{rms,Ne}}=\sqrt{\frac{M_{Ne}}{M_{H_2}}}=\sqrt{\frac{20}{2}}=\sqrt{10}:1\]

Two containers of equal volumes contain H₂ and O₂ at the same temperature. If the number of molecules of these gases is also equal, then the ratio of the pressures exerted by them will be

1. 1 : 1
2. 4 : 1
3. 8 : 1
4. 16 : 1

Answer: (a) 1 : 1. The pressure of the gas is \[P=\frac13\frac{nM}{V}v_{rms}^2=\frac13\frac{nM}{V}\cdot\frac{3RT}{M}=\frac{nRT}{V}\] Given \(n_{H_2}=n_{O_2}\), \(T_{H_2}=T_{O_2}\) and \(V_{H_2}=V_{O_2}\), so \(P_{H_2}=P_{O_2}\).

The density of carbon dioxide gas at 0 °C and at a pressure of \(1.0\times10^5\) newton/metre² is 1.98 kg/m³. Find the root mean square velocity of its molecules at 0 °C. The pressure is constant.

1. 39 m/s
2. 3.09 m/s
3. 389 m/s
4. 38.9 m/s

Answer: (c) 389 m/s. We know \(P=\dfrac13\rho v_{rms}^2\), so \[v_{rms}=\left(\frac{3P}{\rho}\right)^{1/2}=\sqrt{\frac{3\times1.0\times10^5}{1.98}}\approx389\text{ m/s}\]

6. Concept of absolute temperature

1. For a given gas, \(v_{rms}\propto\sqrt T\). \[P=\frac13\frac{nM}{V}v_{rms}^2\;\Rightarrow\;v_{rms}^2=\frac{3PV}{nM}\] Since \(PV=nRT\), \[v_{rms}^2=\frac{3RT}{M}\;\text{ or }\;v_{rms}=\sqrt{\frac{3PV}{nM}}\] (\(n\) = number of moles).
2. It is clear that at \(T=0\) K, \(v_{rms}=0\), i.e. the rms speed of the molecules of a gas is zero at 0 K. A temperature lower than 0 K is not possible, because then the velocity would become imaginary. So the temperature at which the rms speed of gas molecules is zero is called absolute zero or zero kelvin.

7. Degrees of freedom

The number of modes in which a gas molecule can store energy is called its degrees of freedom. It can also be described as the number of modes in which the motion of a gas molecule takes place.

8. Maxwell's law of equipartition of energy

1. According to this law, for a system in thermal equilibrium all its energy is divided equally among all possible degrees of freedom.
2. At temperature \(T\) K, each degree of freedom contributes on average \(\tfrac12kT\) of energy per molecule and \(\tfrac12RT\) of energy per mole.
3. If \(f\) is the number of degrees of freedom, the mean energy of a molecule at temperature \(T\) is \(\bar E=\dfrac{fkT}{2}\) and the total energy per mole is \(\dfrac{fRT}{2}\).
4. Total internal energy \(U\) = potential energy + kinetic energy. For an ideal gas the potential energy is 0, so \(U\) = kinetic energy \(=\dfrac{fRT}{2}\) (per mole).
5. For the translatory motion of any gas \(f=3\), so the mean translatory KE per molecule \(=\tfrac32kT\), the mean translatory KE per mole \(=\tfrac32RT\), and the mean translatory KE per gram \(=\tfrac32\dfrac{RT}{M}\).
6. As we already have \(P=\tfrac23E\), we get \(E=\tfrac32P\), where \(E\) is the energy per unit volume. So the mean translatory KE in unit volume of a gas is \(\tfrac32P\), where \(P\) is the pressure of the gas.
7. Total energy \(E_0=\tfrac32PV\) (for volume \(V\)).

9. Different kinetic energies of a gas

1. Mean molar kinetic energy (\(E\)): the translational kinetic energy of one mole of a gas (\(6.023\times10^{23}\) molecules). \[E=\tfrac12Mv_{rms}^2=\tfrac32RT\]
2. Mean translational KE of one molecule \(=\tfrac32kT\).
3. If \(M\) is the molecular weight in kg, i.e. the mass of one mole of the gas, then the mean KE of one gram of gas \(=\dfrac32\dfrac{RT}{M}=\dfrac32\dfrac{kT}{m}\), where \(m\) = mass of each molecule in kg.

Examples based on different kinetic energies of a gas

At what temperature will the root mean square velocity of oxygen molecules be sufficient to escape from the earth?

1. \(1.6\times10^5\) K
2. \(16\times10^5\) K
3. \(0.16\times10^5\) K
4. \(160\times10^5\) K

Answer: (a) \(1.6\times10^5\) K. Equating \(\dfrac32kT=\dfrac12mv_e^2\), where \(v_e=11.1\) km/s is the escape velocity of the earth and \(m=5.32\times10^{-26}\) kg is the mass of one oxygen molecule, \[T=\frac{mv_e^2}{3k}=\frac{5.32\times10^{-26}\times(11.1\times10^3)^2}{3\times1.38\times10^{-23}}\approx1.6\times10^5\text{ K}\]

The first excited state of the hydrogen atom is 10.2 eV above its ground state. What temperature is needed to excite hydrogen atoms to the first excited level?

1. \(7.88\times10^4\) K
2. \(0.788\times10^4\) K
3. \(78.8\times10^4\) K
4. \(788\times10^4\) K

Answer: (a) \(7.88\times10^4\) K. The KE per atom is \(\tfrac32kT\), and the required KE is 10.2 eV \(=10.2\times1.6\times10^{-19}\) J. \[T=\frac23\times\frac{\text{K.E. per atom}}{k}=\frac23\times\frac{10.2\times1.6\times10^{-19}}{1.38\times10^{-23}}\approx7.88\times10^{4}\text{ K}\] where \(k=1.38\times10^{-23}\) J/mol-K (J/K per molecule).

At what temperature does the average translational kinetic energy of a molecule in a gas equal the kinetic energy of an electron accelerated from rest through a potential difference of 5 volt?

1. \(386.5\times10^3\) K
2. \(3.865\times10^3\) K
3. \(0.38\times10^3\) K
4. \(38.65\times10^3\) K

Answer: (d) \(38.65\times10^3\) K. The KE of the electron is 5 eV \(=5\times1.6\times10^{-19}\) J. Equating it to \(\tfrac32kT\): \[T=\frac{2\times5\times1.6\times10^{-19}}{3\times1.38\times10^{-23}}\approx38.65\times10^3\text{ K}\]

10. Difference between ideal gas and real gas

11. Gas laws

1. Boyle's law: at a given temperature, the pressure of a given mass of gas is inversely proportional to its volume. That is, \(P\propto\dfrac1V\) (\(T\) = constant), or \(PV\) = constant. Graphically:

2. Charles' law: at a given pressure, the volume of a given mass of gas is proportional to its absolute temperature. That is, \(V\propto T\) (\(P\) = constant), with the temperature in kelvin. Graphically:

3. Gay-Lussac's law: at a given volume, the pressure of a given mass of gas is proportional to its absolute temperature. That is, \(P\propto T\) (\(V\) = constant), or \(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\). Graphically:

4. Ideal gas equation: using all the laws above, an equation for \(n\) moles of an ideal gas can be written in the form \[\mathbf{PV=nRT}\;\text{ also }\;PV=NkT\] where \(n\) = number of moles, \(N=N_0n\) = number of molecules, \(N_0\) = Avogadro's constant, \(k\) = Boltzmann's constant \(=R/N_0\). The number of moles \(=\dfrac mM\), where \(m\) = total mass of the gas and \(M\) = molecular weight. \[\Rightarrow\;PV=m\left(\frac RM\right)T=mrT\] where \(r\) = specific gas constant \(=\dfrac RM\).
   Note: if the mass of the gas is also changed, then Boyle's, Charles' and Gay-Lussac's laws cannot be applied. However, we can use the relation \[\frac{P_1V_1}{n_1T_1}=\frac{P_2V_2}{n_2T_2}\]
5. Dalton's law of partial pressure: "The pressure exerted by a mixture of several gases equals the sum of the pressures exerted by each gas occupying the same volume as that of the mixture." For \(n\) gases, \(P=P_1+P_2+P_3+\dots+P_n\), where \(P\) = pressure exerted by the mixture and \(P_1,P_2,P_3,\dots,P_n\) = partial pressures of the component gases.
6. Avogadro's law: at the same temperature and pressure, equal volumes of all gases contain an equal number of molecules or moles.
   Important facts: (a) at \(T=273\) K and \(P=1\) atm (STP), 22.4 litre of each gas has \(N_0=6.023\times10^{23}\) molecules; in other words, one mole of gas at STP occupies 22.4 litre of volume. (b) This means the mass of 22.4 litre of a gas at STP, in grams, is equal to its molecular weight. (c) This implies that 32 g oxygen (O₂), 28 g nitrogen (N₂), 2 g hydrogen (H₂), etc. occupy the same volume, 22.4 litres, at STP.
7. Graham's law of diffusion: when two gases at the same pressure and temperature are allowed to diffuse into each other, the rate of diffusion of each gas is inversely proportional to the square root of the density of the gas. That is, \(D\propto\dfrac1{\sqrt d}\), where \(D\) = rate of diffusion and \(d\) = density of the gas. \[\Rightarrow\;\frac{D_1}{D_2}=\sqrt{\frac{d_2}{d_1}}\]

Examples based on Boyle's law

A sample of oxygen with volume 1000 cc at a pressure of 1 atm is compressed to a volume of 900 cc. What pressure is needed to do this if the temperature is kept constant?

Answer: The temperature is constant, so \(P_1V_1=P_2V_2\). \[P_2=P_1\frac{V_1}{V_2}=(1\text{ atm})\times\frac{1000}{900}=1.11\text{ atm}\]

Examples based on Charles' law

1500 ml of a gas at a room temperature of 23 °C is inhaled by a person whose body temperature is 37 °C. If the pressure stays constant, what will be the volume of the gas in the lungs of the person?

Answer: The pressure and the amount of gas are constant, therefore \(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\). \[V_2=V_1\times\frac{T_2}{T_1}\] Here \(T_1=273+23=296\) K and \(T_2=273+37=310\) K. \[V_2=1500\times\frac{310}{296}\approx1570.95\text{ ml}\]

Examples based on Gay-Lussac's law

A sample of helium gas is at a pressure of 1 atm when the volume is 100 ml and its temperature is 27 °C. What will be the temperature of the gas if the pressure becomes 2 atm and the volume remains 100 ml?

Answer: The volume remains constant, so \(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\). \[T_2=\frac{P_2}{P_1}\times T_1=\frac21\times300\;[T_1=273+27=300\text{ K}]=600\text{ K}=600-273=327\,^\circ\text{C}\]

Examples based on Dalton's law of partial pressure

A vessel of volume \(2\times10^{-3}\) m³ contains 0.1 mol of hydrogen gas and 0.2 mol of helium. If the temperature of the mixture is 300 K, calculate the pressure due to the component gases and the mixture.

Answer: According to Dalton's law, \(P_{mixture}=P_H+P_{He}\). \[P_{mixture}=\frac{RT}{V}(n_H+n_{He})=\frac{8.31\times300}{2\times10^{-3}}(0.1+0.2)\approx3.74\times10^{5}\text{ N/m}^2\] (hydrogen contributes \(1.25\times10^{5}\) N/m² and helium \(2.49\times10^{5}\) N/m²).

Examples based on the ideal gas equation

The quantity \(\dfrac{PV}{kT}\) represents

1. Mass of the gas
2. Kinetic energy of the gas
3. Number of moles of the gas
4. Number of molecules in the gas

Answer: (d) Number of molecules in the gas. By the ideal gas equation \(PV=nRT\), and \(R=N_Ak\), so \(PV=nN_AkT\). With \(N=nN_A\) as the number of molecules, \(\dfrac{PV}{kT}=N\).

A sample of argon gas is at a pressure of 1 atm. When the volume is 100 mL the temperature is 35 °C. What must be its temperature if the pressure becomes 720 mm of Hg and its volume 200 mL?

Answer: Here \(PV=RT\), or \(\dfrac{PV}{T}=R\) (constant), so \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\). Given \(P_1=1\) atm = 760 mm of Hg, \(V_1=100\) mL, \(T_1=35\,^\circ\text{C}=308\) K, \(P_2=720\) mm of Hg, \(V_2=200\) mL. \[\frac{760\times100}{308}=\frac{720\times200}{T_2}\;\Rightarrow\;T_2=\frac{720\times200\times308}{760\times100}\approx584\text{ K}=584-273=311\,^\circ\text{C}\]

12. Mean free path

1. The path traversed in a straight line by a molecule of a gas before it collides with another molecule is called the free path.
2. The average distance travelled between two successive collisions by a molecule is called the mean free path. It is represented by \(\bar\lambda\).
3. Mean free path \(=\dfrac{\text{total distance travelled}}{\text{total number of collisions}}\)

4. If a molecule travels distances \(\lambda_1,\lambda_2,\lambda_3,\dots,\lambda_N\) in \(N\) different collisions during time \(t\), the mean free path will be \[\bar\lambda=\frac{\lambda_1+\lambda_2+\lambda_3+\dots+\lambda_N}{N}\]
5. If the average velocity of a molecule is \(\bar v\), then \[\bar\lambda=\frac{\bar v\,t}{N}\]
6. From kinetic theory the mean free path is \[\bar\lambda=\frac{1}{\sqrt2\,\pi\sigma^2n}\] where \(\sigma\) is the diameter of a molecule and \(n\) is the number of molecules in a unit volume of the gas.
7. If \(m\) is the mass of a molecule, then \[\bar\lambda=\frac1{\sqrt2}\frac{m}{\pi\sigma^2mn}\] but \(mn=\rho\) = density of the gas, \[\therefore\;\bar\lambda=\frac1{\sqrt2}\frac{m}{\pi\sigma^2\rho}\]
8. The mean free path is related to the temperature and the pressure as \[\bar\lambda=\frac{kT}{\sqrt2\,\pi\sigma^2P}\] where \(T\) = temperature, \(k\) = Boltzmann constant, \(P\) = pressure.
9. Therefore \(\bar\lambda\propto\dfrac1\rho\), i.e. the mean free path is inversely proportional to the density of the gas.
10. For constant volume, and hence a constant number density \(n\) of gas molecules, \(\dfrac PT\) is constant, so that \(\bar\lambda\) will not depend on \(P\) and \(T\). But if the volume of a given mass of gas is allowed to change with \(P\) or \(T\), then \(\bar\lambda\propto T\) at constant pressure and \(\bar\lambda\propto\dfrac1P\) at constant temperature. On increasing the temperature at constant pressure the mean free path increases, and on increasing the pressure at constant temperature the mean free path decreases.

Example based on mean free path

The relation between the mean free path of molecules of a gas and its density is

1. \(\bar\lambda\propto\rho\)
2. \(\bar\lambda\propto\dfrac1\rho\)
3. \(\bar\lambda\propto\sqrt\rho\)
4. \(\bar\lambda\propto\dfrac1{\sqrt\rho}\)

Answer: (b) \(\bar\lambda\propto\dfrac1\rho\). Since \(\bar\lambda=\dfrac{m}{\sqrt2\,\pi\sigma^2\rho}\), the mean free path is inversely proportional to the density.

13. Real gas: van der Waals gas equation

1. Real gases do not obey the ideal gas equation.
2. In order to explain the behaviour of real gases, the following two modifications are made in the ideal gas equation:

1. Non-zero size of molecules: a certain portion of the volume of a gas is covered by the molecules themselves. Therefore the space available for the free motion of the molecules of a gas will be slightly less than the volume \(V\) of the gas. This decreases the volume of the gas, which will be \((V-b)\), where \(b\) is about four times the actual volume occupied by the molecules.
2. Intermolecular force: attractive forces act between the molecules of real gases. Due to this, the molecules do not exert on the wall the force which they would have exerted in the absence of intermolecular force. Therefore the observed pressure \(P\) of the gas will be less than the pressure in the absence of intermolecular force. This decrease in pressure is assumed to be equal to \(\dfrac{a}{V^2}\), where \(a\) is a constant. Therefore the pressure of a real gas will be \[P=\frac{RT}{V-b}-\frac{a}{V^2}\] where \(P\) is the observed pressure.

3. The equation obtained by using the above modifications is called van der Waals' equation or the real gas equation. It is as follows: \[\left(P+\frac{a}{V^2}\right)(V-b)=RT\]
4. Real gases behave like ideal gases at low pressure and high temperature.
5. \(a\) and \(b\) are constants and their values are different for different gases.
6. \(a\) is a constant for a gas. Unit of \(a\) = pressure \(\times\) volume² = N \(\times\) m⁴. Dimensions of \(a=\mathrm{M^1L^5T^{-2}}\).
7. \(b\) is a constant for a gas. Its value is four times the volume occupied by the molecules. Unit of \(b\) is the unit of volume = m³. Dimensions of \(b=\mathrm{M^0L^3T^0}\).
8. The pressure exerted by real gases is less than the pressure exerted by ideal gases.
9. The specific heat of a real gas increases with increase of temperature.

14. Critical point or critical constants

1. Critical temperature (\(T_C\)): the temperature below which a gas can be liquefied only by an increase in pressure, and above which no liquefaction is possible whatever the pressure \(P\) may be.
2. Critical pressure (\(P_C\)): the minimum pressure required to liquefy a gas when it is at the critical temperature (\(T_C\)).
3. Critical (specific) volume (\(V_C\)): the volume of unit mass of a gas at critical pressure and temperature.
4. Triple point: the temperature at which all the three states of matter coexist in equilibrium.

\[T_C=\frac{8a}{27Rb},\qquad P_C=\frac{a}{27b^2},\qquad V_C=3b\]

(where \(a\) and \(b\) are van der Waals constants)

=T o t a l n u m b e ro f m o l e c u l e s/Nu m b e r o f m o l . s
(f) Number of moles =M a s so f g a s/M o l e c u l awrt . o f g a s

or n =m/M

(g) The mass of 1 mol of a gas is equal to its molecular weight.