Ideal gas | Conceptual notes | Numerical problem solutions | Short question answer discussion | Grade 11 (Physics) | Physics in Depth

Kinetic theory of gases

• The kinetic theory of gases relates the macroscopic properties of gases (e.g., pressure , temperature ) to the microscopic properties of gas molecules (e.g., speed, kinetic energy ).



• The kinetic theory of gases attempts to develop a model of the molecular behaviour of an ideal gas.



• Assumptions of kinetic theory of gases :
  
  
  • All gases are made up of molecules moving randomly in all directions.
  
  
  
  • The size of a molecule is much smaller than the average separation between the molecules.
  
  
  
  • The molecules exert no force on each other or on the walls of the container except during collision.
  
  
  
  • All collisions between two molecules or between a molecule and a wall are perfectly elastic. Also, the time spent during a collision is negligibly small.
  
  
  
  • The molecules obey Newton's laws of motion.
  
  
  
  • When a gas is left for sufficient time, it comes to a steady state.
  
  
  
• Pressure exerted by the gas on the walls of a container :
  \[P=\frac{1}{3} \rho \bar{c}^2... (i)\] \(\rho\) is the density of gas and \(\bar{c}^2\) is the mean square speed of the gas.
  It can also be written as, \[PV=\frac{1}{3} M \bar{c}^2... (ii)\] Here, M is the toal mass of gas taken and \(M=nm\); n is the total number of molecules in the sample and m is the mass of each gas molecule.
  So, equation (ii) now becomes, \[PV=\frac{1}{3} n m \bar{c}^2 ... (iii)\]



• Average translational kinetic energy of a gas molecule :
  \[\text{K.E.}=\frac{3}{2}kT\] Here, \(k=\frac{R}{N}\) is the boltzmann constant ( N is the number of molecules in 1 mole of a gas i.e., Avogadro's number).



• Total random translational kinetic energy of the gas molecule :
  \[\text{K.E.}=\frac{3}{2}nRT\] Here, n is the number of mole of a gas molecule.



• Root mean square speed :
  The square root of mean square speed is called root - mean square speed or rms speed . \[c_{rms}=\sqrt{\bar{c}^2}\] It can be written into different forms: \[c_{rms}=\sqrt{\frac{3P}{\rho}} ... (i)\] \[c_{rms}=\sqrt{\frac{3RT}{M}} ... (ii)\] \[c_{rms}=\sqrt{\frac{3kT}{m}} ... (iii)\] \[\therefore c_{rms} \propto \sqrt{T}\]

Short Answer Questions Solution

1. When we place a gas cylinder on a van and the van moves, does the kinetic energy of the molecules increase? Does the temperature increase?
   Solution:
   It depends on whether the van is moving with uniform velocity or non - uniform velocity. If the van is moving with uniform velocity, the velocity of the gases with respect to the cylinder does not change and hence their kinetic energies does not change. However, if the motion of van is not uniform (i.e., either accelerating or deccelerating), their velocities change which will change their kinetic energy. Change in kinetic energy increases the temperature of gas. \[\text{K.E.} \propto T\]



2. Do you expect the gas in a cooking gas cylinder to obey the ideal gas equation?
   Solution:
   For the gas to behave as an ideal gas, it should be at low pressure and high temperature. In contrast to this, the gas in a cooking gas cylinder is under high pressure and at room temperature. Another reason is that, the gas in a cooking gas cylinder is in liquid state, and liquid state does not behave as an ideal gas.



3. Is it possible to boil water at room temperature, say 30\(^0\) C?
   Solution:
   Yes, it is. Boiling point reduces with increase in the pressure. If we reduce the pressure so that the boiling point is reduced from 100\(^\circ\) C to 30\(^\circ\) C, the water starts boiling.



4. Which has more atoms a kilogram of hydrogen or a kilogram of iron?
   Solution:
   One mole of any substance contains 6\(\times\) 10\(^{23}\) number of atoms.
   So, 1 gram of hydrogen gas contains 6\(\times\) 10\(^{23}\) number of atoms.
   1 kg of hydrogen gas contains 6\(\times\) 10\(^{23}\) \(\times\) 1000 atoms = 6 \(\times\) 10\(^{26}\) atoms.
   
   Also,
   
   56 g of iron has 6\(\times\) 10\(^{23}\) atoms.
   1 g of iron has \(\frac{6 \times 10^{23}\times 1000}{56}\) atoms = 0.108 \(\times\) 10\(^{26}\) atoms.
   Hence, a kilogram of hydrogen has more atoms than a kilogram of iron.



5. If the absolute temperature of the gas is doubled, what happens to the root mean square speed of the molecules?
   Solution:
   The root mean square speed of the moleules is directly proportional to the absolute temperature of the gas. i.e., \[c_{rms} \propto \sqrt{T}\] So, when the absolute temperature is doubled, \[c_{rms}' \propto \sqrt{2T}\] Comparing this two, \[c_{rms}' = \sqrt{2}\times c_{rms}\] So, on doubling the absolute temperature, the root mean square speed of the molecules increases by \(\sqrt{2}\) times.



6. Is average kinetic theory of gases directly proportional to the absolute temperature? Explain.
   Solution:
   From kinetic theory of gas, the average kinetic energy per molecule of the gas is \[\frac{1}{2}m\bar{c}^2=\frac{3}{2}kT\] Where \textit{m} is the mass of a molecule and \(\bar{c}^2\) is the mean square speed of the molecules of the gas.
   So, from above expression, average kinetic energy of a gas is directly proportional to the absolute temperature. As the temperature is increased, the average kinetic energy of the molecules increases.



7. Air escaping out from an air hose at a gas station always feel cool. Why?
   Solution:
   The pressure exerted by a gas is given by, \[P=\frac{1}{3} \rho \bar{c}^2 = \frac{2}{3}\frac{1}{2} \rho \bar{c}^2 = \frac{2}{3} \hspace{0.1cm}\text{K.E.}\] Where K.E. = \(\frac{1}{2}\rho \bar{c}^2\) is the kinetic energy per unit volume of a gas.
   From this relation, it is clear that pressure of a gas is directly proportional to the kinetic energy per volume of the gas. As the pressure decreases, K.E. decreases and hence temperature also decreases. Pressure outside the air hose is less than pressure inside it and due to this, temperature inside becomes more than outside. Thus, the air escaping out from air hose at a gas station always feel cold.

Numerical Problems Solution

1. Find the rms speed of oxygen molecules in a gas at 300 K.
   Solution:
   
   mass of oxygen molcule, m = 32\(\times\)1.66\(\times\)10\(^{-27}\) kg
   Boltzmann's constant, k = 1.38\(\times\)10\(^{-23}\) J/mol/K
   absolute temperature, T = 300 K
   rms speed, \(c_{rms}\) = ?
   
   From kinetic theory of gases, \[\begin{align*} \frac{1}{2}m \bar{c}^2&=\frac{3}{2}kT\\ \bar{c}^2&=\frac{3kT}{m}\\ c_{rms}=\sqrt{\bar{c^2}}&=\sqrt{\frac{3kT}{m}}\\ &=\sqrt{\frac{3\times 1.38 \times 10^{-23}\times 300}{32\times 1.66\times 10^{-27}}}\\ \therefore c_{rms}&=483.54 \hspace{0.1cm}\text{m/s} \end{align*}\]



2. Calculate the root mean square speed at 0\(^\circ\) C of i) hydrogen molecules ii) oxygen molecules, assuming 1 mole of a gas occupies a volume of 2\(\times 10^{-2}\) m\(^3\) at 0\(^\circ\) C and 10\(^5\) Nm\(^{-2}\) pressure. (Relative molecular mass of hydrogen and oxygen =2 and 32 respectively.)
   Solution:
   
   Absolute temperature, T = 273 K
   Pressure , P = 10\(^5\) Nm\(^{-2}\)
   Molar volume, V = 2\(\times 10^{-2}\) m\(^3\)
   
   For hydrogen molecules,
   
   molar mass, M = 2 g = 2\(\times\)10\(^{-3}\) kg
   root mean square speed, \(c_{rms}\) = ?
   
   From kinetic theory of gas,
   \[\begin{align*} P&=\frac{1}{3}\rho \bar{c}^2\\ \bar{c}^2&=\frac{3P}{\rho}\\ c_{rms}=\sqrt{\bar{c}^2}&=\sqrt{\frac{3PV}{M}}\\ c_{rms}&=\sqrt{\frac{3 \times 10^5 \times 2 \times 10^{-2}}{2\times 10^{-3}}}\\ \therefore c_{rms}&=1732.05 \hspace{0.1cm}\text{m/s} \end{align*}\]
   For oxygen molecules,
   
   molar mass, M = 32 g = 32\(\times\)10\(^{-3}\) kg
   root mean square speed, \(c_{rms}\) = ?
   
   \[\begin{align*} c_{rms}&=\sqrt{\frac{3PV}{M}}\\ &=\sqrt{\frac{3\times 10^5 \times 2\times 10^2}{32\times 10^{-3}}}\\ \therefore c_{rms}&=433.01 \hspace{0.1cm}\text{m/s} \end{align*}\]



3. Calculate the root mean square speed and the average kinetic energy of a molecule of oxygen gas at a temperature of 300 K.
   Mass of oxygen molecule = 32\(\times\)1.66\(\times\)10\(^-{27}\) kg, universal gas constant = 8.31 J mole\(^{-1}\)K\(^{-1}\), Avogadro's number= 6\(\times\)10\(^23\) molecules mole\(^{-1}\). ( Refer to q.no. 1. Answers : 483.54 m/s and 6.23\(\times\) 10\(^{-21}\) J)
   



4. Air at 273 K and 1.01\(\times\)10\(^5\) Nm\(^{-2}\) pressure contains 2.70\(\times\)10\(^{25}\) molecules per cubic meter. How many molecules per cubic metre will there be at a place where the temperature is 223 K and the pressure is 1.33\(\times\)10\(^4\) Nm\(^{-2}\).
   Solution:
   initial pressure, \(P_1\) = 1.01 \(\times\) 10\(^5\) N m\(^{-2}\)
   initial temperature, \(T_1\) = 273 K
   final pressure, \(P_2\) = 1.33\(\times\)10\(^4\) Nm\(^{-2}\)
   final temperature, \(T_2\) = 223 K
   initial no. of molecules per cubic meter, \(n_1\) = 2.70\(\times\)10\(^{25}\)
   final no. of molecules per cubic meter, \(n_2\) = ?
   
   From kinetic theory of gases,
   
   \[\begin{align*} P&=\frac{1}{3}\rho \bar{c}^2\\ &=\frac{1}{3}\frac{M}{V}\bar{c}^2\\ &=\frac{1}{3}\frac{Nm}{V}\bar{c}^2\\ &=\frac{1}{3}nm\bar{c}^2 \end{align*}\] Here, M is the total mass of gas, N is the no. of molecules and m is the mass of each molecule.
   So, from this relations, we can write,
   \[P_1=\frac{1}{3}n_1 m \bar{c_1}^2 ... (i)\] \[P_2=\frac{1}{3}n_2 m \bar{c_2}^2 ... (ii)\] Dividing eqn (ii) by eqn (i), we get,
   \[\begin{align*} \frac{P_2}{P_1}&=\frac{n_2}{n_1}\times \frac{\bar{c_2}^2}{\bar{c_1}^2}\\ \frac{P_2}{P_1}&=\frac{n_2}{n_1}\times \frac{T_2}{T_1}\hspace{0.1cm}\because \bar{c}^2 \propto T\\ n_2&=\frac{P_2}{P_1}\times \frac{T_2}{T_1}\times n_1\\ &=\frac{1.33 \times 10^4}{1.01 \times 10^5}\times \frac{273}{223}\times 2.70 \times 10^{25}\\ \therefore n_2&=4.35 \times 10^{16} \hspace{0.1cm}\text{molecules/m}^3 \end{align*}\]



5. A vessel of volume 2000 cm\(^3\) contains 0.1 mole of oxygen and 0.2 mole of carbon dioxide. If the temperature of the mixture is 300 K, find its pressure.
   Solution:
   
   no. of moles of oxygen, \(n_1\) = 0.1
   np. of moles of carbon dioxide , \(n_2\) = 0.2
   temperature of the mixture, T = 300 K
   volume, V = 2000 cm\(^3\) = 2 \(\times\) 10\(^{-3}\) m\(^3\)
   universal gravitational constant, R = 8.314 J/mol/K
   pressure of the mixture, P = ?
   
   We have from ideal gas equation, \[P=\frac{nRT}{V}\] Let, \(P_1\) and \(P_2\) be the pressure exerted by oxygen and carbon dioxide respectively.
   Now,
   \[\begin{align*} P_1&=\frac{n_1RT}{V}\\ &=\frac{0.1 \times 8.314 \times 300}{2\times 10^{-3}}\\ &=124710 \hspace{0.1cm} \text{Pa} \end{align*}\]
   Then,
   
   \[\begin{align*} P_2&=\frac{n_2 R T}{V}\\ &=\frac{0.2 \times 8.314 \times 300}{2\times 10^{-3}}\\ &=249420 \hspace{0.1cm} \text{Pa} \end{align*}\] So, the pressure of the mixture is,
   
   \[P_1+P_2=124710+249420=374130 \hspace{0.1cm} \text{Pa}\]



6. The average translational kinetic energy of air molecules is 0.040eV (1 eV = 1.6\(\times\)10\(^{-19}\) J). Calculate the temperature of the air. Boltzmann constant k = 1.38\(\times\)10\(^{-23}\) J/K.
   Solution:
   
   K.E. of air molecules, K.E. = 0.040eV = 6.4 \(\times\) 10\(^{-21}\) J
   Boltzmann constant, k = 1.38\(\times\)10\(^{-23}\) J/K
   temperature of air, T = ?
   
   \[\begin{align*} \text{K.E.}&=\frac{3}{2}kT\\ 6.4 \times 10^{-21}&=\frac{3}{2}\times 1.38 \times 10^{-23} \times T\\ T&=\frac{6.4 \times 10^{-21}\times 2}{3\times 1.38 \times 10^{-23}}\\ \therefore T&=309.18 \hspace{0.1cm}\text{K} \end{align*}\]



7. The density of an ideal gas is 1.25\(\times\)10\(^{-3}\) g/c.c. at STP. Calculate the molecular weight of the gas.
   Solution:
   
   density of gas, \(\rho\) = 1.25\(\times\)10\(^{-3}\) g/c.c. = 1.25 kg/m\(^3\)
   temperature, T = 273 K
   
   pressure, P = 10\(^5\) Pa
   molecular weight of the gas, M = ?
   
   From ideal gas equation,
   \[\begin{align*} PV&=nRT\\ PV&=\frac{m}{M}RT\\ P&=\frac{m}{M}\frac{RT}{V}\\ P&=\frac{\rho RT}{M}\\ M&=\frac{\rho RT}{P}\\ &=\frac{1.25 \times 8.314 \times 273}{10^5}\\ \therefore M&=0.028 \hspace{0.1cm}\text{kg} \end{align*}\]



8. Find the rms speed of argon atoms at 323 K.(Same as q.no. 1)



9. The speeds of 11 molecules are 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0 Km/s. What is their rms speed?
   Solution:
   
   \(c_{rms}=\sqrt{\bar{c}^2}\)
   
   Now, \[\begin{align*} \bar{c}^2&=\frac{2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2}{11}\\ &=59\hspace{0.1cm} \text{km/s} \end{align*}\]
   Then,
   \[\begin{align*} c_{rms}&=\sqrt{\bar{c}^2}\\ &=\sqrt{59}\\ \therefore c_{rms}&=7.68 \hspace{0.1cm}\text{km/s} \end{align*}\]



10. What is the total random translational kinetic energy of the molecules in 1 mole of gas at a temperature of 27\(^\circ\) C?
    Solution:
    
    number of moles, n = 1
    temperature , T = 300 K
    universal molar gas constant, R = 8.314 J/mol/K
    
    Total random translational kinetic energy is, \[\begin{align*} &=\frac{3}{2}nRT\\ &=\frac{3}{2}\times 1 \times 8.314 \times 300\\ &=3741.3 \hspace{0.1cm}\text{J} \end{align*}\]