Unit 6 - Gaseous State

Learning Objectives

After studying this unit, students will be able to

• State the laws governing an ideal gas
• Demonstrate gas laws in various real life situations
• Derive ideal gas equation and perform calculations using it
• State the deviations from ideal behavior
• Derive van der Waals equation
• Define Graham’s law of diffusion
• Define compressibility factor
• Explain critical phenomena
• Derive critical constants in terms of van der Waal’s constants
• Explain Andrew’s isotherms of carbon dioxide
• Describe Joule-Thomson effect and liquefaction of gases

6.1 Introduction

We can survive for weeks without food, days without water, but only minutes without air. Thus, we inhale a lungful of air every few seconds, keep some of the molecules for our own end, and some of the molecules that our body no longer needs, and exhale the mixture back into the surrounding air. The air around us is in the gaseous state, which is the simplest of the states of matter. Although the chemical behaviour of gases depends on their composition, all the gases have remarkably similar physical behaviour.

Do you know the difference between gas and vapour?

Gas is a substance that is normally in a gaseous state at room temperature and 1 atm pressure, while vapour is the gaseous form of any substance that is a liquid or solid at room temperature and 1 atm pressure.

Earth is surrounded by an atmosphere of air whose composition in volume percentage is roughly \( 78\% \) nitrogen, \( 21\% \) oxygen and \( 1\% \) other gases. Of the known elements, only eleven are gases under normal atmospheric conditions. The elements hydrogen \( \mathrm{H_2} \), nitrogen \( \mathrm{N_2} \), oxygen \( \mathrm{O_2} \), fluorine \( \mathrm{F_2} \) and chlorine \( \mathrm{Cl_2} \) exist as gaseous diatomic molecules. Another form of oxygen, namely, ozone \( \mathrm{O_3} \) is also a gas at room temperature. The noble gases, namely, helium \( \mathrm{He} \), neon \( \mathrm{Ne} \), argon \( \mathrm{Ar} \), krypton \( \mathrm{Kr} \), xenon \( \mathrm{Xe} \) and radon \( \mathrm{Rn} \) of \( 18^{\mathrm{th}} \) group are monatomic gases. Compounds such as carbon monoxide (CO), carbon dioxide \( \mathrm{CO_2} \), nitrogen dioxide \( \mathrm{NO_2} \) and nitric oxide \( \mathrm{NO} \) also exist in gaseous state under normal atmospheric conditions. In this unit you will learn the properties of gases and other related phenomena.

6.2 The Gas Laws

The gas laws have played a major role in the development of chemistry. The physical properties of all gases are governed by the gas laws that were formulated based on the studies of the properties like pressure, volume, etc., as a function of temperature. Before studying the gas laws in detail, let us understand an important parameter, namely, the pressure.

Pressure is defined as force divided by the area to which the force is applied. The SI unit of pressure is pascal which is defined as 1 Newton per square meter \( \mathrm{N \ m^{-2}} \). There are other units that are commonly used and their relation with the SI unit is as follows.

Table 6.1 Units of pressure

6.2.1 Boyle’s Law: Pressure-Volume Relationship

Robert Boyle performed a series of experiments to study the relation between the pressure and volume of gases. The schematic of the apparatus used by him is shown in figure 6.1.

Mercury was added through the open end of the apparatus such that the mercury level on both ends are equal as shown in the figure 6.1(a). Add more amount of mercury until the volume of the trapped air is reduced to half of its original volume as shown in figure 6.1(b). The pressure exerted on the gas by the addition of excess mercury is given by the difference in mercury levels of the tube. Initially the pressure exerted by the gas is equal to 1 atm as the difference in height of the mercury levels is zero. When the volume is reduced to half, the difference in mercury levels increases to \( 760 \ \mathrm{mm} \). Now the pressure exerted by the gas is equal to 2 atm. It led him to conclude that at a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.

Mathematically, the Boyle’s law can be written as

(\( T \) and \( n \) are fixed, \( T \)- temperature, \( n \)- number of moles)

\( k \) - proportionality constant

When we rearrange equation 6.2

Boyle’s law is applicable to all gases regardless of their chemical identity (provided the pressure is low). Therefore, for a given mass of a gas under two different sets of conditions at constant temperature we can write

The \( PV \) relationship can be understood as follows. The pressure is due to the force of the gas particles on the walls of the container. If a given amount of gas is compressed to half of its volume, the density is doubled and the number of particles hitting the unit area of the container will be doubled. Hence, the pressure would increase twofold.

Consequence of Boyle’s law

The pressure-density relationship can be derived from the Boyle’s law as shown below.

where “m” is the mass, \( \mathrm{d_1} \) and \( \mathrm{d_2} \) are the densities of gases at pressure \( \mathrm{P_1} \) and \( \mathrm{P_2} \).

In other words, the density of a gas is directly proportional to pressure.

Most commercial aeroplanes fly at about 30,000 feet altitude. The pressure decreases with the increase in altitude as there are fewer molecules per unit volume of air. Hence, while at air, the pressure around the aeroplane will be so low that one could pass out for lack of oxygen. For this reason aeroplanes cabins are artificially pressurized.

Similarly, the effect of drop in pressure is felt as a little pain in the ears by a person while ascending a mountain in a plain. Though the external pressure drops, the internal pressure within the ear cavities remains the same. This creates an imbalance. The greater internal pressure forces the eardrum to bulge outward causing pain. The excess air within the ear cavities escapes after some time and with the help of yawning and thereby equalizing the internal and external pressure to relieve the pain.

Underwater divers are advised not to hold the breath unnecessarily while diving. Do you know why?

In figure (6.3) let us find the missing parameters (volume in 6.3(b) and pressure in 6.3(c))

Solution:

According to Boyle’s law, at constant temperature for a given mass of gas at constant temperature,

and

Evaluate Yourself

1. Freon-12, the compound widely used in the refrigerator system as coolant causes depletion of ozone layer. Now it has been replaced by eco-friendly compounds. Consider \( 1.5 \ \mathrm{dm^3} \) sample of gaseous Freon at a pressure of \( 0.3 \ \mathrm{atm} \). If the pressure is changed to \( 1.2 \ \mathrm{atm} \) at a constant temperature, what will be the volume of the gas increased or decreased?

2. Inside a certain automobile engine, the volume of air in a cylinder is \( 0.375 \ \mathrm{dm^3} \), when the pressure is 1.05 atm. When the gas is compressed to a volume of \( 0.125 \ \mathrm{dm^3} \) at the same temperature, what is the pressure of the compressed air?

6.2.2 Charles Law (Volume-Temperature Relationship)

The relationship between volume of a gas and its temperature was examined by J. A. C. Charles. He observed that for a fixed mass of a gas at constant pressure, the volume is directly proportional to its temperature (K). Mathematically it can be represented as (at constant \( P \) and \( n \))

If the temperature of the gas increases, the volume also increases in direct proportion, so that \( \frac{\mathrm{V}}{\mathrm{T}} \) is a constant. For the same system at constant pressure, one can write

For example, if a balloon is moved from an ice cold water bath to a boiling water bath, the temperature of the gas increases. As a result, the gas molecules inside the balloon move faster and gas expands. Hence, the volume increases.

Variation of volume with temperature at constant pressure

The plot of the volume of the gas against its temperature at a given pressure is shown in the figure 6.5. From the graph it is clear that the volume of the gas linearly increases with temperature at a given pressure. Such lines are called isobars. It can be expressed by the following straight line equation.

where \( \mathrm{T} \) is the temperature in degree Celsius and m & C are constants.

When \( \mathrm{T} = 0^{\circ}\mathrm{C} \) the volume becomes \( \mathrm{V_0} \). Hence, \( \mathrm{V_0} = \mathrm{C} \) and slope of the straight line m is equal to \( \Delta \mathrm{V} / \Delta \mathrm{T} \). Therefore the above equation can be written in the following form.

(\( n, P \) are constant)

Divide the equation 6.7 by \( \mathrm{V_0} \)

Charles and Gay Lussac found that under constant pressure, the relative increase in volume per degree increase in temperature is same for all gases. The relative increase in volume per \( ^{\circ}\mathrm{C} \) ( \( \alpha \) ) is equal to \( \frac{1}{V_0}\left(\frac{\Delta V}{\Delta T}\right) \)

Therefore

Charles found that the coefficient of expansion is approximately equal to \( 1/273 \). It means that at constant pressure for a given mass, for each degree rise in temperature, all gases expand by \( 1/273 \) of their volume at \( 0^{\circ}\mathrm{C} \).

If we extrapolate the straight line in the figure 6.5 beyond the experimental measurements, the straight line intersects the temperature axis (x-axis) at \( -273^{\circ}\mathrm{C} \). This shows that the volume of the gas becomes zero at \( -273^{\circ}\mathrm{C} \); more precisely this temperature is \( -273.15^{\circ}\mathrm{C} \). Beyond this temperature the gas would have a negative volume which is physically impossible. For this reason, this temperature was defined as absolute zero by Kelvin and he proposed a new temperature scale with absolute zero as starting point which is now called Kelvin scale. The only difference between the Kelvin scale of temperature and Celsius scale of temperature is that the zero position is shifted. The boiling and freezing point of water in both scales are given below.

Example:

In figure 6.6 let us find the missing parameters (volume in 6.6(b) and temperature in 6.6(c))

Solution:

According to Charles law,

Evaluate Yourself

3. A sample of gas has a volume of \( 3.8 \ \mathrm{dm^3} \) at an unknown temperature. When the sample is submerged in ice water at \( 0^{\circ}\mathrm{C} \) its volume gets reduced to \( 2.27 \ \mathrm{dm^3} \). What is its initial temperature?

6.2.3 Gay-Lussac’s Law (Pressure-Temperature Relationship)

Joseph Gay-Lussac stated that, at constant volume the pressure of a fixed mass of a gas is directly proportional to temperature.

If \( \mathrm{P_1} \) and \( \mathrm{P_2} \) are the pressures at temperatures \( \mathrm{T_1} \) and \( \mathrm{T_2} \) respectively, then from Gay Lussac’s law

Activity - 1

The table below contains the values of pressure measured at different temperatures for 1 mole of an ideal gas. Plot the values in a graph and verify the Gay Lussac’s law. [Lines in the pressure vs temperature graph are known as isochores (constant volume) of a gas.]

6.2.4 Avogadro’s Hypothesis

Avogadro hypothesised that equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules. The mathematical form of Avogadro’s hypothesis may be expressed as

where \( \mathrm{V_1} \) & \( \mathrm{n_1} \) are the volume and number of moles of a gas and \( \mathrm{V_2} \) & \( \mathrm{n_2} \) are a different set of values of volume and number of moles of the same gas at same temperature and pressure.

Evaluate Yourself

4. An athlete in a kinesiology research study has his lung volume of \( 7.05 \ \mathrm{dm^3} \) during a deep inhalation. At this volume the lungs contain 0.312 mole of air. During exhalation the volume of his lung decreases to \( 2.35 \ \mathrm{dm^3} \). How many moles of air does the athlete exhale during exhalation? (assume pressure and temperature remain constant)

6.3 Ideal Gas Equation

The gaseous state is described completely using the following four variables \( \mathrm{T}, \mathrm{P}, \mathrm{V} \) and \( \mathrm{n} \) and their relationships were governed by the gas laws studied so far.

We can combine these equations into the following general equation that describes the physical behaviour of all gases.

where \( \mathrm{R} \) is the proportionality constant called universal gas constant.

The above equation can be rearranged to give the ideal gas equation

We already know that pressure is expressed in many different units (Table 6.1) hence it is important to know the values of gas constant R in different units as well.

We can calculate R using the equation,

For conditions in which P is 1 atm, volume \( 22.414 \ \mathrm{dm^3} \) for 1 mole at 273.15 K.

Under standard conditions (STP) where \( \mathrm{P} = 1 \ \mathrm{bar} \) (\( 10^5 \) pascal), \( \mathrm{V} = 22.71 \times 10^{-3} \ \mathrm{m^3} \) for 1 mole of a gas at 273.15 K

The ideal gas equation is a relationship between four variables (P, V, T, n). Since it describes the state of any gas, it is referred to as the equation of state of gases.

Let us calculate the pressure exerted by 2 moles of sulphur hexafluoride in a steel vessel of volume \( 6 \ \mathrm{dm^3} \) at \( 70^{\circ}\mathrm{C} \) assuming it is an ideal gas.

We will use the ideal gas equation for this calculation as below:

Evaluate Yourself

5. A small bubble rises from the bottom of a lake where the temperature and pressure are \( 8^{\circ}\mathrm{C} \) and \( 6.4 \ \mathrm{atm} \) to the water surface, where the temperature is \( 25^{\circ}\mathrm{C} \) and pressure is 1 atm. Calculate the final volume in (mL) of the bubble, if its initial volume is \( 2.1 \ \mathrm{mL} \).

6.4 Mixture of Gases - Dalton’s Law of Partial Pressures

Studies of non-reacting gaseous mixtures showed that in a gaseous mixture each component behaves independently. For a gaseous mixture, it is important to know, how the pressure of individual component contributes to the total pressure of the mixture.

John Dalton stated that “the total pressure of a mixture of non-reacting gases is the sum of partial pressures of the gases present in the mixture” where the partial pressure of a component gas is the pressure that it would exert if it were present alone in the same volume and temperature. This is known as Dalton’s law of partial pressures.

i.e., for a mixture containing three gases 1, 2 and 3 with partial pressures \( p_1 \), \( p_2 \) and \( p_3 \) in a container with volume V, the total pressure \( P_{\text{total}} \) will be given by

Assuming that the gases behave ideally,

The partial pressure can also be expressed as

where \( x_i \) is the mole fraction of the \( i^{th} \) component.

Application of Dalton’s law

In a reaction involving the collection of gas by downward displacement of water, the pressure of dry gas collected can be calculated using Dalton’s law.

\( P_{\text{water vapour}} \) is generally referred as aqueous tension and its values are available for air at various temperatures.

Let us understand Dalton’s law by solving this problem. A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is 2 atm at a fixed temperature.

Solution:

Evaluate Yourself

6. (a) A mixture of He and \( \mathrm{O_2} \) were used in the ‘air’ tanks of underwater divers for deep dives. For a particular dive \( 12 \ \mathrm{dm^3} \) of \( \mathrm{O_2} \) at \( 1 \ \mathrm{atm} \) and \( 46 \ \mathrm{dm^3} \) of He at \( 1 \ \mathrm{atm} \) were both pumped into a \( 5 \ \mathrm{dm^3} \) tank. Calculate the partial pressure of each gas and the total pressure in the tank at \( 298 \ \mathrm{K} \).

(b) A sample of solid \( \mathrm{KClO_3} \) (potassium chlorate) was heated in a test tube to obtain \( \mathrm{O_2} \) according to the reaction

The oxygen gas was collected by downward displacement of water at \( 295 \ \mathrm{K} \). The total pressure of the mixture is \( 772 \ \mathrm{mm} \) of Hg. The vapour pressure of water is \( 26.7 \ \mathrm{mm} \) of Hg at \( 300 \ \mathrm{K} \). What is the partial pressure of the oxygen gas?

6.4.1 Graham’s Law of Diffusion

Gases have a tendency to occupy all the available space. When two non-reactive gases are allowed to mix, the gas molecules migrate from region of higher concentration to a region of lower concentration. This property of gas which involves the movement of the gas molecules through another gases is called diffusion. Effusion is another process in which a gas escapes from a container through a very small hole.

The rate of diffusion or effusion is inversely proportional to the square root of molar mass. This statement is called Graham’s law of diffusion/effusion.

Mathematically,

Otherwise

When diffusing gases are at different pressures \( (\mathrm{P_A}, \mathrm{P_B}) \)

where \( r_{\mathrm{A}} \) and \( r_{\mathrm{B}} \) are the rates of diffusion of A and B and \( M_{\mathrm{A}} \) and \( M_{\mathrm{B}} \) are their respective molar masses.

Example:

1. An unknown gas diffuses at a rate of 0.5 time that of nitrogen at the same temperature and pressure. Calculate the molar mass of the unknown gas.

Solution:

Squaring on both sides

Evaluate Yourself

7. A flammable hydrocarbon gas of particular volume is found to diffuse through a small hole in 1.5 minutes. Under the same conditions of temperature and pressure an equal volume of bromine vapour takes 4.73 min to diffuse through the same hole. Calculate the molar mass of the unknown gas and suggest what this gas might be. (Given that molar mass of bromine \( = 159.8 \ \mathrm{g / mol} \))

6.5 Deviation from Ideal Gas Behaviour

The kinetic theory of gases (postulates of kinetic theory of gases are described in physics text book) which is the basis for the gas equation \( \mathrm{PV = nRT} \) assumes that the individual gas molecules occupy negligible volume when compared to the total volume of the gas and there is no attractive force between the gas molecules. Gases whose behaviour is consistent with these assumptions under all conditions are called ideal gases. But in practice both these assumptions are not valid under all conditions. For example, the fact that gases can be liquefied shows that the attractive force exists among molecules. Hence, there is no gas which behaves ideally under all conditions. The non-ideal gases are called real gases. The real gases tend to approach the ideal behaviour under certain conditions.

6.5.1 Compressibility Factor Z

The deviation of real gases from ideal behaviour is measured in terms of a ratio of \( \mathrm{PV} \) to \( \mathrm{nRT} \). This is termed as compressibility factor. Mathematically,

For ideal gases \( \mathrm{PV = nRT} \), hence the compressibility factor, \( Z = 1 \) at all temperatures and pressures. For these gases the plot of \( Z \) vs \( P \) should be a straight line parallel to the pressure axis. When a gas deviates from ideal behaviour, its \( Z \) value deviates from unity. For all gases, at very low pressures and very high temperature the compressibility factor approaches unity and they tend to behave ideally. The plot of the compressibility factor vs pressure for some common gases are shown in Figure 6.8.

When the pressure is low, the volume of the container is very large compared to the volume of the gas molecules so that individual volume of the gas molecules can be neglected. In addition, the molecules in a gas are far apart and attractive forces are negligible. As the pressure increases, the density of gas also increases and the molecules are much closer to one another. Hence, the intermolecular force becomes significant enough to affect the motion of the molecules and the gas will not behave ideally.

At high temperatures the average kinetic energy of the molecules is very high and hence intermolecular attractions will become insignificant. As the temperature decreases, the average kinetic energy of molecules also decreases, hence the molecular attraction is enhanced. The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point. The Boyle point varies with the nature of the gas. Above the Boyle point, for real gases, \( Z > 1 \), i.e., the real gases show positive deviation. Below the Boyle point, the real gases first show a decrease for \( Z \), reaches a minimum and then increases with the increase in pressure. So, it is clear that at low pressure and at high temperature, the real gases behave as ideal gases.

6.5.2 Compressibility Factor for Real Gases

The compressibility factor \( Z \) for real gases can be rewritten

Substituting 6.18 in 6.17

Where \( V_{\text{real}} \) is the molar volume of the real gas and \( V_{\text{ideal}} \) is the molar volume of it when it behaves ideally.

6.5.3 Van der Waals Equation

J. D. Van der Waals made the first mathematical analysis of real gases. His treatment provides us an interpretation of real gas behaviour at the molecular level. He modified the ideal gas equation \( \mathrm{PV = nRT} \) by introducing two correction factors, namely, pressure correction and volume correction.

Pressure Correction

The pressure of a gas is directly proportional to the force created by the bombardment of molecules on the walls of the container. The speed of a molecule moving towards the wall of the container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is lower than the ideal pressure of the gas. Hence, van der Waals introduced a correction term to this effect.

Van der Waals found out the forces of attraction experienced by a molecule near the wall are directly proportional to the square of the density of the gas.

where n is the number of moles of gas and V is the volume of the container

where a is proportionality constant and depends on the nature of gas

Therefore,

Volume Correction

As every individual molecule of a gas occupies a certain volume, the actual volume is less than the volume of the container, V. Van der Waals introduced a correction factor \( V' \) to this effect. Let us calculate the correction term by considering gas molecules as spheres.

where \( v_m \) is the volume of a single molecule

Excluded volume for single molecule

Excluded volume for n molecule

Where b is van der Waals constant which is equal to \( 4V_m \)

Replacing the corrected pressure and volume in the ideal gas equation \( \mathrm{PV = nRT} \) we get the van der Waals equation of state for real gases as below,

The constants \( a \) and \( b \) are van der Waals constants and their values vary with the nature of the gas. It is an approximate formula for the non-ideal gas.

6.6 Pressure-Volume Isotherms of Carbon Dioxide

Andrew’s Isotherm

Thomas Andrew gave the first complete data on pressure-volume-temperature of a substance in the gaseous and liquid states. He plotted isotherms of carbon dioxide at different temperatures which is shown in Figure 6.12. From the plots we can infer the following.

At low temperature isotherms, for example, at \( 13^{\circ}\mathrm{C} \) as the pressure increases, the volume decreases along AB and is a gas until the point B is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist and the pressure remains constant. At C, the gas is completely converted into liquid. If the pressure is higher than at C, only the liquid is compressed so, there is no significant change in the volume. The successive isotherms shows similar trend with the shorter flat region. i.e., the volume range in which the liquid and gas coexist becomes shorter. At the temperature of \( 31.1^{\circ}\mathrm{C} \) the length of the shorter portion is reduced to zero at point P. In other words, the \( \mathrm{CO_2} \) gas is liquefied completely at this point. This temperature is known as the liquefaction temperature or critical temperature of \( \mathrm{CO_2} \). At this point the pressure is 73 atm. Above this temperature \( \mathrm{CO_2} \) remains as a gas at all pressure values. It is then proved that many real gases behave in a similar manner to carbon dioxide.

Though the nature of isotherm remains similar, the critical temperature, the corresponding pressure and volume are characteristics of a particular gas.

Now we can define the critical constants as follows.

• Critical temperature \( (T_c) \) of a gas is defined as the temperature above which it cannot be liquefied even at high pressure.
• Critical pressure \( (P_c) \) of a gas is defined as the minimum pressure required to liquefy 1 mole of a gas at its critical temperature.
• Critical volume \( (V_c) \) is defined as the volume occupied by 1 mole of a gas at its critical temperature and critical pressure.

The critical constants of some common gases are given in Table 6.2.

Table 6.2 Critical constants of some gases

Evaluate Yourself

8. Critical temperature of \( \mathrm{H_2O} \), \( \mathrm{NH_3} \), and \( \mathrm{CO_2} \) are 647.4, 405.5 and 304.2 K, respectively. When we start cooling from a temperature of 700 K which will liquefy first and which will liquefy finally?

6.6.1 Derivation of Critical Constants from van der Waals Constants

The van der Waals equation for n moles is

For 1 mole

From the equation we can derive the values of critical constants \( P_c \), \( V_c \) and \( T_c \) in terms of a and b, the van der Waals constants. On expanding the above equation

Multiply equation (6.24) by \( \frac{V^2}{P} \)

When the above equation is rearranged in powers of V

The equation (6.26) is a cubic equation in V. On solving this equation, we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume \( V_c \). The pressure and temperature become \( P_c \) and \( T_c \) respectively

i.e., \( V = V_c \)

As equation (6.26) is identical with equation (6.27), we can equate the coefficients of \( V^2 \), V and constant terms in (6.26) and (6.27).

Divide equation (6.30) by equation (6.29)

When equation (6.31) is substituted in (6.29)

Substituting the values of \( V_c \) and \( P_c \) in equation (6.28),

The critical constants can be calculated using the values of van der Waals constant of a gas and vice versa.

6.7 Liquefaction of Gases

For important commercial operations such as LPG and rocket fuels, we require gases in their liquid state. The liquefaction methods are based on the Joule-Thomson effect. He observed appreciable cooling when the compressed gas is forced through an orifice plug into a low-pressure region. This phenomenon of lowering of temperature when a gas is made to expand adiabatically from a region of high pressure into a region of low pressure is known as Joule-Thomson effect. This effect is observed only below a certain temperature, which is a characteristic one for each gas. This temperature below which a gas obeys Joule-Thomson effect is called inversion temperature \( (T_i) \). This value is given using van der Waals constants \( a \) and \( b \).

Gases like \( \mathrm{O_2} \), He, \( \mathrm{N_2} \) and \( \mathrm{H_2} \) have very low \( T_c \), hence Joule-Thomson effect can be applied for cooling effectively. At the inversion temperature, no rise or fall in temperature of a gas occurs while expanding. But above the inversion temperature, the gas gets heated up when allowed to expand through a hole.

There are different methods used for liquefaction of gases:

1. In Linde’s method, Joule-Thomson effect is used to get liquid air or any other gas.
2. In Claude’s process, the gas is allowed to perform mechanical work in addition to Joule-Thomson effect so that more cooling is produced.
3. In Adiabatic process, cooling is produced by removing the magnetic property of magnetic material such as gadolinium sulphate. By this method, a temperature of \( 10^{-4} \ \mathrm{K} \) i.e., as low as \( 0 \ \mathrm{K} \) can be achieved.

Summary

The state of a gas is defined by a relationship between the four independent variables pressure (P), volume (V), temperature (T) and number of moles (n). The relationship between these parameters is governed by different gas laws as summarised below.

Gases that obey the equation \( \mathrm{PV = nRT} \) under all conditions are called ideal gases. But in practice there is no ideal gas. Gases tend to behave ideally at high temperatures and at low pressures. For real gases, van der Waals modified the ideal gas equation as

Critical temperature \( (T_c) \) of a gas is defined as the temperature above which it cannot be liquefied at any pressure. Critical pressure \( (P_c) \) of a gas is defined as the minimum pressure required to liquefy 1 mole of a gas at its critical temperature. Critical volume \( (V_c) \) is defined as the volume occupied by 1 mole of a gas at its critical temperature and critical pressure.

The critical constants are related to Van der Waals constants as follows

When a gas is made to expand adiabatically from a region of high pressure into a region of low pressure, the temperature of the gas is reduced rapidly and this is known as Joule-Thomson effect. This effect is used in the liquefaction of gases.

Questions

I. Choose the best answer

1. Gases deviate from ideal behavior at high pressure. Which of the following statement(s) is correct for non-ideality? a) at high pressure the collision between the gas molecule become enormous b) at high pressure the gas molecules move only in one direction c) at high pressure, the volume of gas become insignificant d) at high pressure the intermolecular interactions become significant

2. Rate of diffusion of a gas is a) directly proportional to its density b) directly proportional to its molecular weight c) directly proportional to its square root of its molecular weight d) inversely proportional to the square root of its molecular weight

3. Which of the following is the correct expression for the equation of state of van der Waals gas? a) \( \left(P + \frac{a}{n^2V^2}\right)(V - nb) = nRT \) b) \( \left(P + \frac{na}{n^2V^2}\right)(V - nb) = nRT \) c) \( \left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT \) d) \( \left(P + \frac{n^2a^2}{V^2}\right)(V - nb) = nRT \)

4. When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules a) are above inversion temperature b) exert no attractive forces on each other c) do work equal to the loss in kinetic energy d) collide without loss of energy

5. Equal weights of methane and oxygen are mixed in an empty container at 298 K. The fraction of total pressure exerted by oxygen is a) \( \frac{1}{3} \) b) \( \frac{1}{2} \) c) \( \frac{2}{3} \) d) \( \frac{1}{3} \times \frac{273}{298} \)

6. The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called a) Critical temperature b) Boyle temperature c) Inversion temperature d) Reduced temperature

7. In a closed room of \( 1000 \ \mathrm{m^3} \) a perfume bottle is opened up. The room develops a smell. This is due to which property of gases? a) Viscosity b) Density c) Diffusion d) None

8. A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be a) At the center of the tube b) Near the hydrogen chloride bottle c) Near the ammonia bottle d) Throughout the length of the tube

9. The value of universal gas constant depends upon a) Temperature of the gas b) Volume of the gas c) Number of moles of the gas d) units of Pressure and volume

10. The value of the gas constant R is a) \( 0.082 \ \mathrm{dm^3 \ atm} \) b) \( 0.987 \ \mathrm{cal \ mol^{-1} \ K^{-1}} \) c) \( 8.3 \ \mathrm{J \ mol^{-1} \ K^{-1}} \) d) \( 8 \ \mathrm{erg \ mol^{-1} \ K^{-1}} \)

11. Use of hot air balloon in sports and meteorological observation is an application of a) Boyle’s law b) Newton’s law c) Kelvin’s law d) Brown’s law

12. The table indicates the value of van der Waals constant ‘a’ in \( (\mathrm{dm^3})^2 \ \mathrm{atm \ mol^{-2}} \)

The gas which can be most easily liquefied is a) \( \mathrm{O_2} \) b) \( \mathrm{N_2} \) c) \( \mathrm{NH_3} \) d) \( \mathrm{CH_4} \)

13. Consider the following statements i) Atmospheric pressure is less at the top of a mountain than at sea level ii) Gases are much more compressible than solids or liquids iii) When the atmospheric pressure increases the height of the mercury column rises Select the correct statement a) I and II b) II and III c) I and III d) I, II and III

14. Compressibility factor for \( \mathrm{CO_2} \) at \( 400 \ \mathrm{K} \) and \( 71.0 \ \mathrm{bar} \) is \( 0.8697 \). The molar volume of \( \mathrm{CO_2} \) under these conditions is a) \( 22.04 \ \mathrm{dm^3} \) b) \( 2.24 \ \mathrm{dm^3} \) c) \( 0.41 \ \mathrm{dm^3} \) d) \( 19.5 \ \mathrm{dm^3} \)

15. If temperature and volume of an ideal gas is increased to twice its values, the initial pressure P becomes a) 4P b) 2P c) P d) 3P

16. At identical temperature and pressure, the rate of diffusion of hydrogen gas is \( 3\sqrt{3} \) times that of a hydrocarbon having molecular formula \( \mathrm{C_nH_{2n-2}} \). What is the value of n? a) 8 b) 4 c) 3 d) 1

17. Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape. (NEET phase I) a) \( \frac{3}{8} \) b) \( \frac{1}{2} \) c) \( \frac{1}{8} \) d) \( \frac{1}{4} \)

18. The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion i.e., \( \alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p \). For an ideal gas \( \alpha \) is equal to a) T b) \( \frac{1}{T} \) c) P d) none of these

19. Four gases P, Q, R and S have almost same values of ‘b’ but their ‘a’ values (a, b are Vander Waals Constants) are in the order \( Q < R < S < P \). At a particular temperature, among the four gases the most easily liquefiable one is a) P b) Q c) R d) S

20. Maximum deviation from ideal gas is expected from (NEET) a) \( \mathrm{CH_4(g)} \) b) \( \mathrm{NH_3(g)} \) c) \( \mathrm{H_2(g)} \) d) \( \mathrm{N_2(g)} \)

21. The units of Vander Waals constants ‘b’ and ‘a’ respectively a) \( \mathrm{mol \ L^{-1}} \) and \( \mathrm{L \ atm^2 \ mol^{-1}} \) b) \( \mathrm{mol \ L} \) and \( \mathrm{L \ atm \ mol^2} \) c) \( \mathrm{mol^{-1} \ L} \) and \( \mathrm{L^2 \ atm \ mol^{-2}} \) d) none of these

22. Assertion: Critical temperature of \( \mathrm{CO_2} \) is 304 K, it can be liquefied above 304 K. Reason: For a given mass of gas, volume is directly proportional to pressure at constant temperature a) both assertion and reason are true and reason is the correct explanation of assertion b) both assertion and reason are true but reason is not the correct explanation of assertion c) assertion is true but reason is false d) assertion is false but reason is true

23. What is the density of \( \mathrm{N_2} \) gas at \( 227^{\circ}\mathrm{C} \) and \( 5.00 \ \mathrm{atm} \) pressure? \( \mathrm{R} = 0.082 \ \mathrm{L \ atm \ K^{-1} \ mol^{-1}} \) a) \( 1.40 \ \mathrm{g/L} \) b) \( 2.81 \ \mathrm{g/L} \) c) \( 3.41 \ \mathrm{g/L} \) d) \( 0.29 \ \mathrm{g/L} \)

24. Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas? (T is measured in K)

25. \( 25 \ \mathrm{g} \) of each of the following gases are taken at \( 27^{\circ}\mathrm{C} \) and \( 600 \ \mathrm{mmHg} \) pressure. Which of these will have the least volume? a) HBr b) HCl c) HF d) HI

II. Answer the following questions

26. State Boyle’s law.

27. Name two items that can serve as a model for Gay Lussac’s law and explain.

28. Give the mathematical expression that relates gas volume and moles.

29. What are ideal gases? In what way real gases differ from ideal gases.

30. Can a Van der Waals gas with \( a = 0 \) be liquefied? explain.

31. Suppose there is a tiny sticky area on the wall of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?

32. Explain the following observations a) Aerated water bottles are kept under water during summer b) Liquid ammonia bottle is cooled before opening the seal c) The tyre of an automobile is inflated to slightly lesser pressure in summer than in winter d) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude

33. Give suitable explanation for the following facts about gases. a) Gases don’t settle at the bottom of a container b) Gases diffuse through all the space available to them

34. Suggest why there is no hydrogen \( \mathrm{(H_2)} \) in our atmosphere. Why does the moon have no atmosphere?

35. Explain whether a gas approaches ideal behavior or deviates from ideal behaviour if a) it is compressed to a smaller volume at constant temperature. b) the temperature is raised while keeping the volume constant c) more gas is introduced into the same volume and at the same temperature

36. Which of the following gases would you expect to deviate from ideal behaviour under conditions of low temperature \( \mathrm{F_2} \), \( \mathrm{Cl_2} \) or \( \mathrm{Br_2} \)? Explain.

37. Distinguish between diffusion and effusion.

38. Aerosol cans carry clear warning of heating of the can. Why?

39. Would it be easier to drink water with a straw on the top of Mount Everest?

40. Write the Van der Waals equation for a real gas. Explain the correction term for pressure and volume.

41. Derive the values of critical constants in terms of van der Waals constants.

42. Why do astronauts have to wear protective suits when they are on the surface of moon?

43. When ammonia combines with HCl, \( \mathrm{NH_4Cl} \) is formed as white dense fumes. Why do more fumes appear near HCl?

44. A sample of gas at \( 15^{\circ}\mathrm{C} \) at 1 atm has a volume of \( 2.58 \ \mathrm{dm^3} \). When the temperature is raised to \( 38^{\circ}\mathrm{C} \) at 1 atm does the volume of the gas increase? If so, calculate the final volume.

45. A sample of gas has a volume of \( 8.5 \ \mathrm{dm^3} \) at an unknown temperature. When the sample is submerged in ice water at \( 0^{\circ}\mathrm{C} \), its volume gets reduced to \( 6.37 \ \mathrm{dm^3} \). What is its initial temperature?

46. Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen in a vessel of volume \( 37.6 \ \mathrm{dm^3} \) at \( 298 \ \mathrm{K} \), and the sample B is in a vessel of volume \( 16.5 \ \mathrm{dm^3} \) at \( 298 \ \mathrm{K} \). Calculate the number of moles in sample B.

47. Sulphur hexafluoride is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas in a steel vessel of volume \( 5.43 \ \mathrm{dm^3} \) at \( 69.5^{\circ}\mathrm{C} \), assuming ideal gas behaviour.

48. Argon is an inert gas used in light bulbs to retard the vaporization of the tungsten filament. A certain light bulb containing argon at 1.2 atm and \( 18^{\circ}\mathrm{C} \) is heated to \( 85^{\circ}\mathrm{C} \) at constant volume. Calculate its final pressure in atm.

49. A small bubble rises from the bottom of a lake where the temperature and pressure are \( 6^{\circ}\mathrm{C} \) and 4 atm to the water surface, where the temperature is \( 25^{\circ}\mathrm{C} \) and pressure is 1 atm. Calculate the final volume in (mL) of the bubble, if its initial volume is \( 1.5 \ \mathrm{mL} \).

50. Hydrochloric acid is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of \( 154.4 \times 10^{-3} \ \mathrm{dm^3} \) of a gas at a pressure of \( 742 \ \mathrm{mm} \) of Hg at a temperature of \( 298 \ \mathrm{K} \). What mass of hydrogen gas (in mg) did the student collect?

51. It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for \( \mathrm{N_2} \) gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas?

52. A tank contains a mixture of \( 52.5 \ \mathrm{g} \) of oxygen and \( 65.1 \ \mathrm{g} \) of \( \mathrm{CO_2} \) at \( 300 \ \mathrm{K} \). The total pressure in the tank is 9.21 atm. Calculate the partial pressure (in atm) of each gas in the mixture.

53. A combustible gas is stored in a metal tank at a pressure of 2.98 atm at \( 25^{\circ}\mathrm{C} \). The tank can withstand a maximum pressure of 12 atm after which it will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal \( = 1100 \ \mathrm{K} \)).

Summary of Gas Laws

1. Boyle’s Law - \( P \propto \frac{1}{V} \) (\( T, n \) constant)
2. Charles Law - \( V \propto T \) (\( P, n \) constant)
3. Gay Lussac - \( P \propto T \) (\( V, n \) constant)
4. Avogadro’s Hypothesis - \( V \propto n \) (\( T, P \) constant)
5. Ideal gas equation - \( PV = nRT \)

ICT Corner

Properties of Gases

By using this tool you will be able to understand the relationship between T, P and V of gases and verify the gas laws.

Please go to the URL https://phet.colorado.edu/en/simulation/gas-properties (or) Scan the QR code.

Step-1: Open the Browser and type the URL given (or) Scan the QR Code. In the website click the gas properties applet, then you will see a Java applet named ‘gas properties’ opening up as shown in the figure.

Set up instructions:

• In the chamber you can pump some gas molecules by using the pump handle.
• The temperature and pressure of the gas molecules in the chamber can be monitored by using the corresponding gauges.
• The volume of the chamber can be altered by moving the left side wall.
• The temperature of the chamber can be altered by heating or cooling using the slider.
• Any of the three properties can be kept as constant by making appropriate selections in the box.

Verifying the gas laws:

By using the above instructions you can understand the gas laws. To understand the Boyle’s law, keep the temperature constant and pump few molecules of gas and keep it as constant. Now reduce the volume by pushing the left side wall then you can see that the pressure of the gas increases. Similarly the other two gas laws can also be verified.