Pressure Exerted By A Gas

    9.2.1 Expression for pressure exerted by a gas#

    Consider a monoatomic gas of N molecules each having a mass m inside a cubical container of side \( l \) as shown in the Figure 9.1 (a).

    Figure 9.1 (a) Container of gas molecules

    Figure 9.1 (b) Collision of a molecule with the wall

    The molecules of the gas are in random motion. They collide with each other and also with the walls of the container. As the collisions are elastic in nature, there is no loss of kinetic energy, but a change in momentum occurs.

    The molecules of the gas exert pressure on the walls of the container due to collision on it. During each collision, the molecules impart certain momentum to the wall. Due to transfer of momentum, the walls experience a continuous force. The force experienced per unit area of the walls of the container determines the pressure exerted by the gas. It is essential to determine the total momentum transferred by the molecules in a short interval of time.

    A molecule of mass m moving with a velocity \( \bar{\nu} \) having components \( (v_{x}, v_{y}, v_{z}) \) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. This is shown in the Figure 9.1 (b). The components of velocity of the molecule after collision are \( (- \nu_{x}, \nu_{y}, \nu_{z}) \).

    The x-component of momentum of the molecule before collision \( = m\nu_{x} \)

    The x-component of momentum of the molecule after collision \( = - mv_{x} \)

    The change in momentum of the molecule in x direction

    \( = \) Final momentum - initial momentum \( = - mv_{x} - mv_{x} = - 2mv_{x} \)

    According to law of conservation of linear momentum, the change in momentum of the wall \( = 2mv_{x} \)

    Figure 9.2 Number of molecules hitting the wall

    The number of molecules hitting the right side wall in a small interval of time \( \Delta t \) is calculated as follows.

    The molecules within the distance of \( v_{x}\Delta t \) from the right side wall and moving towards the right will hit the wall in the time interval \( \Delta t \). This is shown in the Figure 9.2. The number of molecules that will hit the right side wall in a time interval \( \Delta t \) is equal to the product of volume \( (Av_{x}\Delta t) \) and number density of the molecules \( (n) \). Here \( A \) is area of the wall and \( n \) is number of molecules per unit volume \( \left(\frac{N}{V}\right) \). We have assumed that the number density is the same throughout the cube.

    Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side.

    The number of molecules that hit the right side wall in a time interval \( \Delta t \)

    \[ = \frac{n}{2} A v_{x} \Delta t \]

    In the same interval of time \( \Delta t \), the total momentum transferred by the molecules

    \[ \Delta p = \frac{n}{2} A v_{x} \Delta t \times 2 m v_{x} = A v_{x}^{2} m n \Delta t \]

    From Newton’s second law, the change in momentum in a small interval of time gives rise to force.

    The force exerted by the molecules on the wall (in magnitude)

    \[ \mathrm{F} = \frac{\Delta p}{\Delta t} = n m A v_{x}^{2} \]

    Pressure, \( \mathrm{P} = \) force divided by the area of the wall

    \[ \mathrm{P} = \frac{F}{A} = n m v_{x}^{2} \]

    Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term \( v_{x}^{2} \) by the average \( \overline{v_{x}^{2}} \) in the above equation.

    \[ P = n m \overline{v_{x}^{2}} \]

    Since the gas is assumed to move in random direction, it has no preferred direction of motion (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So, \( \overline{v_{x}^{2}} = \overline{v_{y}^{2}} = \overline{v_{z}^{2}} \). The mean square speed is written as

    \[ \overline{v^{2}} = \overline{v_{x}^{2}} + \overline{v_{y}^{2}} + \overline{v_{z}^{2}} = 3 \overline{v_{x}^{2}} \]

    \[ \overline{v_{x}^{2}} = \frac{1}{3} \overline{v^{2}} \]

    Using this, we get

    \[ P = \frac{1}{3} n m \overline{v^{2}} \quad \text{or} \quad P = \frac{1}{3} \frac{N}{V} m \overline{v^{2}} \quad \left[ n = \frac{N}{V} \right] \]

    The following inference can be made from the above equation. The pressure exerted by the molecules depends on

    (i) Number density \( n = \frac{N}{V} \). It implies that if the number density increases then pressure will increase. For example when we pump air inside the cycle tyre or car tyre essentially the number density increases and as a result the pressure increases.

    (ii) Mass of the molecule: Since the pressure arises due to momentum transfer to the wall, larger mass will have larger momentum for a fixed speed. As a result the pressure will increase.

    (iii) Mean square speed: For a fixed mass if we increase the speed, the average speed will also increase. As a result the pressure will increase.

    For simplicity the cubical container is taken into consideration. The above result is true for any shape of the container as the area A does not appear in the final expression. Hence the pressure exerted by gas molecules on the wall is independent of area of the wall (A).

    9.2.2 Kinetic interpretation of temperature#

    To understand the microscopic origin of temperature, rewrite the equation

    \[ P = \frac{1}{3} \frac{N}{V} m \overline{v^{2}} \]

    \[ PV = \frac{1}{3} N m \overline{v^{2}} \]

    Comparing this with ideal gas equation \( PV = NkT \)

    \[ NkT = \frac{1}{3} N m \overline{v^{2}} \]

    \[ kT = \frac{1}{3} m \overline{v^{2}} \]

    Multiply the above equation by \( 3/2 \) on both sides,

    \[ \frac{3}{2} kT = \frac{1}{2} m \overline{v^{2}} \]

    R.H.S of the equation is called average kinetic energy of a single molecule \( (\overline{KE}) \).

    The average kinetic energy per molecule

    \[ \overline{KE} = \epsilon = \frac{3}{2} kT \]

    This implies that the temperature of a gas is a measure of the average translational kinetic energy per molecule of the gas.

    Equation is a very important result from kinetic theory of gas. We can infer the following from this equation.

    (i) The average kinetic energy of the molecule is directly proportional to absolute temperature of the gas. The equation gives the connection between the macroscopic world (temperature) to microscopic world (motion of molecules).

    (ii) The average kinetic energy of each molecule depends only on temperature of the gas not on mass of the molecule. In other words, if the temperature of an ideal gas is measured using thermometer, the average kinetic energy of each molecule can be calculated without seeing the molecule through naked eye.

    By multiplying the total number of gas molecules with average kinetic energy of each molecule, the internal energy of the gas is obtained.

    Internal energy of ideal gas \( U = N \left( \frac{1}{2} m \overline{v^{2}} \right) \)

    By using the above equation

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

    From this equation, we understand that the internal energy of an ideal gas depends only on absolute temperature and is independent of pressure and volume.

    EXAMPLE 9.1

    A football at \( 27^{\circ}C \) has 0.5 mole of air molecules. Calculate the internal energy of air in the ball.

    Solution

    The internal energy of ideal gas \( = \frac{3}{2} NkT \)

    The number of air molecules is given in terms of number of moles so, rewrite the expression as follows

    \[ U = \frac{3}{2} \mu RT \]

    Since \( Nk = \mu R \). Here \( \mu \) is number of moles.

    Gas constant \( R = 8.31 \, \frac{J}{mol \cdot K} \)

    Temperature \( T = 273 + 27 = 300 \, K \)

    \[ U = \frac{3}{2} \times 0.5 \times 8.31 \times 300 = 1869.75 \, J \]

    This is approximately equivalent to the kinetic energy of a man of \( 57 \, \text{kg} \) running with a speed of \( 8 \, \text{m} \, \text{s}^{-1} \).

    9.2.3 Relation between pressure and mean kinetic energy#

    From earlier section, the internal energy of the gas is given by

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

    The above equation can also be written as

    \[ U = \frac{3}{2} PV \quad (\text{since } PV = NkT) \]

    \[ P = \frac{2}{3} \frac{U}{V} = \frac{2}{3} u \]

    From the equation, we can state that the pressure of the gas is equal to two thirds of internal energy per unit volume or internal energy density \( \left(u = \frac{U}{V}\right) \).

    Writing pressure in terms of mean kinetic energy density using the earlier equation

    \[ P = \frac{1}{3} n m \overline{v^{2}} = \frac{1}{3} \rho \overline{v^{2}} \]

    where \( \rho = n m = \) mass density (Note n is number density)

    Multiply and divide R.H.S by 2, we get

    \[ P = \frac{2}{3} \left( \frac{1}{2} \rho \overline{v^{2}} \right) = \frac{2}{3} \overline{KE} \]

    9.2.4 Some elementary deductions from kinetic theory of gases#

    Boyle’s law:

    From earlier, we know that \( PV = \frac{2}{3} U \). But the internal energy of an ideal gas is equal to N times the average kinetic energy \( (\epsilon) \) of each molecule.

    \[ U = N\epsilon \]

    For a fixed temperature, the average translational kinetic energy \( \epsilon \) will remain constant. It implies that

    \[ PV = \frac{2}{3} N\epsilon \quad \text{Thus} \quad PV = \text{constant} \]

    Therefore, pressure of a given gas is inversely proportional to its volume provided the temperature remains constant. This is Boyle’s law.

    Charles’ law:

    From the equation, we get \( PV = \frac{2}{3} U \). For a fixed pressure, the volume of the gas is proportional to internal energy of the gas or average kinetic energy of the gas and the average kinetic energy is directly proportional to absolute temperature. It implies that

    \[ V \propto T \quad \text{or} \quad \frac{V}{T} = \text{constant} \]

    This is Charles’ law.

    Avogadro’s law:

    This law states that at constant temperature and pressure, equal volumes of all gases contain the same number of molecules. For two different gases at the same temperature and pressure, according to kinetic theory of gases,

    From the pressure equation

    \[ P = \frac{1}{3} \frac{N_1}{V} m_1 \overline{v_1^2} = \frac{1}{3} \frac{N_2}{V} m_2 \overline{v_2^2} \]

    where \( \overline{v_1^2} \) and \( \overline{v_2^2} \) are the mean square speed for two gases and \( N_1 \) and \( N_2 \) are the number of gas molecules in two different gases.

    At the same temperature, average kinetic energy per molecule is the same for two gases.

    \[ \frac{1}{2} m_1 \overline{v_1^2} = \frac{1}{2} m_2 \overline{v_2^2} \]

    Dividing the first equation by the second we get \( N_1 = N_2 \).

    This is Avogadro’s law. It is sometimes referred to as Avogadro’s hypothesis or Avogadro’s Principle.

    9.2.5 Root mean square speed \( (v_{rms}) \)#

    Root mean square speed \( (v_{rms}) \) is defined as the square root of the mean of the square of speeds of all molecules. It is denoted by \( v_{\mathrm{rms}} = \sqrt{\overline{v^2}} \).

    From earlier equation can be re-written as,

    \[ \text{mean square speed } \overline{v^2} = \frac{3kT}{m} \]

    root mean square speed,

    \[ v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}} = 1.73 \sqrt{\frac{kT}{m}} \]

    From the equation we infer the following

    (i) rms speed is directly proportional to square root of the temperature and inversely proportional to square root of mass of the molecule. At a given temperature the molecules of lighter mass move faster on an average than the molecules with heavier masses.

    Example: Lighter molecules like hydrogen and helium have high \( v_{rms} \) than heavier molecules such as oxygen and nitrogen at the same temperature.

    (ii) Increasing the temperature will increase the r.m.s speed of molecules.

    We can also write the \( v_{rms} \) in terms of gas constant R. The above equation can be rewritten as follows

    \[ v_{rms} = \sqrt{\frac{3 N_A kT}{N_A m}} \quad \text{Where } N_A \text{ is Avogadro number} \]

    Since \( N_A k = R \) and \( N_A m = M \) (molar mass)

    The root mean square speed or r.m.s speed

    \[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

    The pressure equation can also be written in terms of rms speed \( P = \frac{1}{3} n m v_{rms}^{2} \) since \( v_{rms}^{2} = \overline{v^{2}} \).

    Impact of \( v_{rms} \) in nature:

    1. Moon has no atmosphere. The escape speed of gases on the surface of Moon is much less than the root mean square speeds of gases due to low gravity. Due to this all the gases escape from the surface of the Moon.

    2. No hydrogen in Earth’s atmosphere. As the root mean square speed of hydrogen is much greater than that of nitrogen, it easily escapes from the earth’s atmosphere. In fact, the presence of nonreactive nitrogen instead of highly combustible hydrogen deters many disastrous consequences.

    EXAMPLE 9.2

    A room contains oxygen and hydrogen molecules in the ratio 3:1. The temperature of the room is \( 27^{\circ}C \). The molar mass of \( O_2 \) is \( 32 \, \text{g mol}^{-1} \) and of \( H_2 \) is \( 2 \, \text{g mol}^{-1} \). The value of gas constant R is \( 8.32 \, \text{J mol}^{-1} \text{K}^{-1} \).

    Calculate

    (a) rms speed of oxygen and hydrogen molecule

    (b) Average kinetic energy per oxygen molecule and per hydrogen molecule

    (c) Ratio of average kinetic energy of oxygen molecules and hydrogen molecules

    Solution

    (a) Absolute Temperature \( \mathrm{T} = 27^{\circ} \mathrm{C} = 27 + 273 = 300 \, \mathrm{K} \)

    Gas constant \( \mathrm{R} = 8.32 \, \mathrm{J \, mol^{-1} \, K^{-1}} \)

    For Oxygen molecule: Molar mass \( \mathrm{M} = 32 \, \mathrm{g} = 32 \times 10^{-3} \, \mathrm{kg \, mol^{-1}} \)

    \[ \mathrm{rms \, speed} \, v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.32 \times 300}{32 \times 10^{-3}}} = 483.73 \, \text{m s}^{-1} \approx 484 \, \text{m s}^{-1} \]

    For Hydrogen molecule: Molar mass \( \mathrm{M} = 2 \times 10^{-3} \, \mathrm{kg \, mol^{-1}} \)

    \[ \mathrm{rms \, speed} \, v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.32 \times 300}{2 \times 10^{-3}}} = 1934 \, \text{m s}^{-1} \]

    Note that the rms speed is inversely proportional to \( \sqrt{\mathrm{M}} \) and the molar mass of oxygen is 16 times higher than molar mass of hydrogen. It implies that the rms speed of hydrogen is 4 times greater than rms speed of oxygen at the same temperature. \( \frac{1934}{484} \approx 4 \).

    (b) The average kinetic energy per molecule is \( \frac{3}{2} kT \). It depends only on absolute temperature of the gas and is independent of the nature of molecules. Since both the gas molecules are at the same temperature, they have the same average kinetic energy per molecule. \( k \) is Boltzmann constant.

    \[ \frac{3}{2} kT = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \, \text{J} \]

    (c) Average kinetic energy of total oxygen molecules \( = \frac{3}{2} N_o kT \) where \( N_o \) - number of oxygen molecules in the room.

    Average kinetic energy of total hydrogen molecules \( = \frac{3}{2} N_H kT \) where \( N_H \) - number of hydrogen molecules in the room.

    It is given that the number of oxygen molecules is 3 times more than number of hydrogen molecules in the room. So the ratio of average kinetic energy of oxygen molecules with average kinetic energy of hydrogen molecules is 3:1.

    9.2.6 Mean (or) average speed \( (\overline{v}) \)#

    It is defined as the mean (or) average of all the speeds of molecules.

    If \( \nu_{1}, \nu_{2}, \nu_{3}, \ldots, \nu_{N} \) are the individual speeds of molecules then

    \[ \overline{\nu} = \frac{\nu_{1} + \nu_{2} + \nu_{3} + \ldots + \nu_{n}}{N} = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8kT}{\pi m}} \]

    Here M - Molar Mass and m - mass of the molecule.

    \[ \overline{\nu} = 1.60 \sqrt{\frac{kT}{m}} \]

    9.2.7 Most probable speed \( (V_{mp}) \)#

    It is defined as the speed acquired by most of the molecules of the gas.

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

    \[ \nu_{mp} = 1.41 \sqrt{\frac{kT}{m}} \]

    The derivation of the above equations is beyond the scope of the book.

    Comparison of \( v_{rms}, \overline{\nu} \) and \( \nu_{mp} \)

    Among the speeds \( v_{rms} \) is the largest and \( \nu_{mp} \) is the least

    \[ v_{rms} > \overline{\nu} > \nu_{mp} \]

    Ratio-wise,

    \[ v_{rms} : \overline{\nu} : \nu_{mp} = \sqrt{3} : \sqrt{\frac{8}{\pi}} : \sqrt{2} = 1.732 : 1.6 : 1.414 \]

    EXAMPLE 9.3

    Ten particles are moving at the speed of 2, 3, 4, 5, 5, 5, 6, 6, 7 and \( 9 \, \text{m} \, \text{s}^{-1} \). Calculate rms speed, average speed and most probable speed.

    Solution

    The average speed \( \overline{\nu} = \frac{2 + 3 + 4 + 5 + 5 + 5 + 6 + 6 + 7 + 9}{10} = 5.2 \, \text{m s}^{-1} \)

    To find the rms speed, first calculate the mean square speed \( \overline{\nu^{2}} \)

    \[ \overline{\nu^{2}} = \frac{2^{2} + 3^{2} + 4^{2} + 5^{2} + 5^{2} + 5^{2} + 6^{2} + 6^{2} + 7^{2} + 9^{2}}{10} = 30.6 \, \text{m}^{2} \text{s}^{-2} \]

    The rms speed

    \[ \nu_{rms} = \sqrt{\overline{\nu^{2}}} = \sqrt{30.6} = 5.53 \, \text{m s}^{-1} \]

    The most probable speed is \( 5 \, \text{m s}^{-1} \) because three of the particles have that speed.

    EXAMPLE 9.4

    Calculate the rms speed, average speed and the most probable speed of 1 mole of hydrogen molecules at \( 300 \, \mathrm{K} \). Neglect the mass of electron.

    Solution

    The hydrogen atom has one proton and one electron. The mass of electron is negligible compared to the mass of proton. Mass of one proton \( = 1.67 \times 10^{-27} \, \text{kg} \). One hydrogen molecule \( = 2 \) hydrogen atoms \( = 2 \times 1.67 \times 10^{-27} \, \text{kg} \).

    The average speed

    \[ \bar{\nu} = \sqrt{\frac{8kT}{\pi m}} = 1.60 \sqrt{\frac{kT}{m}} = 1.60 \sqrt{\frac{(1.38 \times 10^{-23}) \times 300}{2(1.67 \times 10^{-27})}} = 1.78 \times 10^{3} \, \text{m s}^{-1} \]

    (Boltzmann Constant \( k = 1.38 \times 10^{-23} \, \text{J K}^{-1} \))

    The rms speed

    \[ \nu_{rms} = \sqrt{\frac{3kT}{m}} = 1.73 \sqrt{\frac{kT}{m}} = 1.73 \sqrt{\frac{(1.38 \times 10^{-23}) \times 300}{2(1.67 \times 10^{-27})}} = 1.93 \times 10^{3} \, \text{m s}^{-1} \]

    Most probable speed

    \[ \nu_{mp} = \sqrt{\frac{2kT}{m}} = 1.41 \sqrt{\frac{kT}{m}} = 1.41 \sqrt{\frac{(1.38 \times 10^{-23}) \times 300}{2(1.67 \times 10^{-27})}} = 1.57 \times 10^{3} \, \text{m s}^{-1} \]

    Note that \( \nu_{rms} > \bar{\nu} > \nu_{mp} \).

    9.2.8 Maxwell-Boltzmann speed distribution function#

    In a classroom, the air molecules are moving in random directions. The speed of each molecule is not the same even though macroscopic parameters like temperature and pressure are fixed. Each molecule collides with every other molecule and they exchange their speed. In the previous section we calculated the rms speed of each molecule and not the speed of each molecule which is rather difficult. In this scenario we can find the number of gas molecules that move with the speed of \( 5 \, \text{m s}^{-1} \) to \( 10 \, \text{m s}^{-1} \) or \( 10 \, \text{m s}^{-1} \) to \( 15 \, \text{m s}^{-1} \) etc. In general our interest is to find how many gas molecules have the range of speed from \( \nu \) to \( \nu + d\nu \). This is given by Maxwell’s speed distribution function.

    \[ N_{\nu} = 4\pi N \left( \frac{m}{2\pi kT} \right)^{\frac{3}{2}} \nu^{2} e^{-\frac{m\nu^{2}}{2kT}} \]

    The above expression is graphically shown as follows

    Figure 9.3 Maxwell’s molecular speed distribution

    From the Figure 9.3, it is clear that, for a given temperature the number of molecules having lower speed increases parabolically \( (\nu^{2}) \) but decreases exponentially \( (e^{-\frac{m\nu^{2}}{2kT}}) \) after reaching most probable speed. The rms speed, average speed and most probable speed are indicated in the Figure 9.3. It can be seen that the rms speed is greatest among the three.