A Tutorial on The Basics of the Gaseous State: Notes, Figures and Problems with Solutions
Target Audience: These notes on Atomic Structure are meant for college freshmen, or high school students in Grades 11 or 12.
might be of use to Indian students preparing for the ISC or CBSE Class
11 and 12 Examinations, IIT JEE (main and advanced), AIEEE; students
from across the world preparing for their A Level Examinations, IB
(International Baccalaureate) or AP Chemistry.
This compilation of notes has been prepared by Anirud Thyagarajan of IIT Kharagpur
The Gaseous State
This topic is mostly theoretical and mathematical in nature.
Matter can be classified into three categories depending upon its physical state namely solid, liquid and gaseous states. Solids have a definite volume and shape; liquids also have a definite volume but no definite shape; gases have neither a definite volume nor a definite shape.
MEASURABLE PROPERTIES OF GASES
Mass, volume, temperature are the important measurable properties of gases.
Mass: The mass of the gas is related to the number of moles as
n = w/M
Where n = number of moles
w = mass of gas in grams
M = molecular mass of the gas
Volume: Since gases occupy the entire space available to them, therefore the gas volume means the volume of the container in which the gas is enclosed.
Units of Volume: Volume is generally expressed in litre or cm3 or dm3 1m3 = 103 litre
= 103 dm3 = 106 cm3.
Pressure: The force exerted by the gas per unit area on the walls of the container is equal to its pressure.
Units of Pressure: The pressure of a gas is expressed in atm, Pa, Nm–2, bar or, lb/In2 (psi).
760 mm = 1 atm = 10132.5 KPa = 101325 Pa = 101325 Nm–2
760 mm of Hg = 1.01325 bar = 1013.25 milli bar = 14.7 lb/2n2 (psi)
Temperature: Temperature is defined as the degree of hotness. The SI unit of temperature is Kelvin. On the Celsius scale water freezes at 0°C and boils at 100°C where as in the Kelvin scale water freezes at 273 K and boils at 373 K.
The state of a sample of gas is defined by 4 variables i.e. P, V, n & T. Gas laws are the simple relationships between any two of these variables when the other two are kept constant.
Boyle’s Law: The changes in the volume of a gas by varying pressure at a constant temperature of a fixed amount of gas was quantified by Robert Boyle in 1662. The law was named after his name as Boyle’s law. It states that:
The volume of a given mass of a gas is inversely proportional to its pressure at a constant temperature.
P ∝ 1/V (n, T constant)
V ∝ 1/P (n, T constant)
i.e. P = K/V (where K is the constant proportionality)
or PV = K (constant)
Let V1 be the volume of a given mass of the gas having pressure P1 at temperature T. Now, if the pressure is changed to P2 at the same temperature, let the volume changes to V2. The quantitative relationship between the four variables P1,V1, P2 and V2 is:
P1V1 = P2V2 (temperature and mass constant)
Graphical Representation of Boyle’s Law
Fig. (a) show the plot of V vs P at a particular temperature. It shows that P increases
Plot (b) shows the plot of PV vs P at particular temperature. It indicates that PV value remains constant inspite of regular increase in P.
DALTON’S LAW OF PARTIAL PRESSURE
The relation between the pressures of the mixture of non-reacting gases enclosed in a vessel to their individual pressure is described in the law. The law was given by John Dalton in 1807. It states that.
At constant temperature, the pressure exerted by a mixture of two or more non-reacting gases enclosed in a definite volume, is equal to the sum of the individual pressures which each gas would exert if present alone in the same volume.
The individual pressures of gases are known as partial pressures.
If P is the total pressure of the mixture of non-reacting gases at temperature T and volume V, and P1, P2, P3 …. represent the partial pressures of the gases, then
P = P1 + P2 + P3+ ……… (T, V are constant)
Partial Pressure in terms of Mole Fraction
Mole fraction c defines the amount of a substance in a mixture as a fraction of total amount of all substances. If n=1 moles of way substance is present in n moles of the mixture, then mole fraction of the substance, X1 = n1 / n
If PN2 is the partial pressure of nitrogen the mixture of SO2 and N2. Then
PN2 = nN2 × RT/V (= no. of moles of N2)
and Pmixture = (nN2 + nSO2) RT / V
dividing we get,
PN2 / Pmixture = (nN2 / nN2 + NSO2) = XN2
or, PN2 = XN2 × Pmixture
GRAHAM’S LAW OF DIFFUSION/EFFUSION
The ability of a gas to spread and occupy the whole available volume irrespective of other gases present in the container is called diffusion.
Effusion is the process by which a gas escapes from one chamber of a vessel through a small opening or an orifice. r = volume diffused / time taken = V / T
Thomas Graham in 1831 proposed the law of gaseous diffusion. The law states under similar conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional to the square roots of their densities.
r ∝ 1 / √d where is the rate of diffusion and d is the density of the gas.
Now, if there are two gases A and B having r1 and r2 as their rates of diffusion and d1 and d2 their densities respectively. Then
r1 ∝ 1 / √d1 and r2 ∝ 1 / √d2 or, √d2/d1 (at same T and P)
r1/r2 = √M2 / 2 M1 / 2 = √M2 / M1 = V1 / V2 T1 / T2
Here M1 and M2 are the molecular masses of the gases having densities d1 and d2 respectively.
Graham’s law of diffusion also holds good for effusion.
Effect of pressure on state of diffusion
The rate of diffusion (r) of a gas at constant temperature is directly preoperational to its pressure
r ∝ P at constatnt temperature
r ∝ 1 / √d at constant temperature
r1 / r2 = P1 / P2 × √M2 / M1
Intermolecular forces or Vander Waal’s forces
Intermolecular forces or vander Waals’ forces originate from the following three types of interactions.
Dipole – Dipole interactions: In case of polar molecules, the vander waals’ forces are mainly due to electrical interaction between oppositively charged ends of molecules (Fig. 1. a) called dipole – dipole interactions. For example, gases such as etc.have permanent dipole moments as a result of which there is appreciable dipole – dipole interactions between the molecules of these gases. The magnitude of this type of interaction depends upon the dipole moment of the molecule concerned. Evidently, greater the dipole moment, stronger is the dipole – dipole interactions. Because of these attractive forces, these gases can be easily liquefied.
Dipole – Induced dipole Interactions: A polar molecule may sometimes polarize a non – polar molecule which lies in its vicinity and thus induces polarity in that molecule just as a magnet induces magnetic polarity in a neutral piece of iron lying close by. The induced dipole then interacts with the dipole moment of the first molecule and thereby the two molecules are attracted together (Fig. 1. b). The magnitude of this interaction, evidently depends upon the magnitude of the dipole moment of the polar molecule and the polarizability of the non – polar molecule. The solubility of inert gases in increases from He to Rn due to a corresponding increase in magnitude of the dipole – induced dipole interactions as the polarizability of the noble gas increases with increase in size from He to Rn.
Momentary dipole – induced dipole interactions: The electrons of neutral molecules keep on oscillating w.r.t. the nuclei of atoms. As a result, at a given instant, one side of the molecule may have a slight excess of electrons relative to the opposite side. Thus a non – polar molecule may become momentarily self – polarized. This polarized molecule may induce a dipole moment in the neighbouring molecule. These two induced dipoles then attract each other (Fig. 1. c). These momentary dipole – induced dipole attractions are also called London forces or dispersive forces. The magnitude of these forces depends upon the following:
(i) Size or molecular mass: The melting points and boiling points of non – polar molecules increase as the size or molecular mass of the molecule increases. For example, the m.p. and b.p. of alkanes, halogens, noble gases etc. increase as the molecular mass of the molecule increases.
(ii) Geometry / Shape : For example, isomer n – pentane has higher boiling point than neo – pentane because the former is zig – zag chain with larger sites of contact and hence large intermolecular forces whereas the latter is nearly spherical and hence has less contact and weaker forces.
KINETIC MOLECULAR THEORY OF GASES
The various gas laws which we have studied so far were based an experimental facts.
The theory that provides an explanation for the various experimental observations about a gas is based on the Kinetic Molecular model. It is assumed that all gases are made up of identical molecules, which are in a state of constant random motion.
Postulates of the Model
The following are the postulates of the model:
• A gas consists of a large number of identical molecules of mass m. The dimensions of these molecules are very small compared to the space between them. Hence the molecules are treated as point masses.
• There are practically no attractive forces between the molecules. The molecules therefore more independently.
• The molecules are in a state of ceaseless and random motion, colliding with each other and with the walls of the container. The direction of their motions changes only on collision. These collisions are known as elastic collisions in which the energy and momenta of the molecules are concerned. In non-elastic collisions these quantities are not conserved.
• The pressure of a gas is the result of collision of molecules with the wall of the container.
• The average kinetic energy of the colliding molecules is directly proportional to its temperature.
Velocities of gas molecules
As per kinetic theory of gases, each molecule is moving with altogether different velocity. Let ‘n’ molecules be present in a given mass of gas, each one moving with velocity v1, v2, v3… vn. The average velocity or UaV = average of all such velocity terms.
Average velocity = u1 + u2 + u2 + ...un / n = n1u1 + n2u2 + n3u3 = n1 + n2 + n3 ...
Uav = √8RT / πM = √8PV / πM = √8P / πb
Root Mean Square Velocity
Maxwell proposed the term Urms as the square root of means of square of all such velocities.
U2rms = u21 + u22 + u23 + ... / n1u21n2u22 + n3u23 + / n1 + n2 + n3
Also Urms = √3RT / M = √3PV / M = √3P / d
Most probable velocity
It is the velocity which is possessed by maximum no. of molecules.
Vmp = √2RT / M = √2PV / M = √2P / d
Furthermore UMP : UAV : Urms : ::: √2RT / M : √8RT / πM : √3RT / M
√2: √8 / π : 1.224
1 : 1.128 : 1.224
Also Uav = Urms × 0.9213
A Few Problems with Solutions
Prob 1. A gas occupies one litre under atmospheric pressure. What will be the volume of the same amount of gas under 750 mm of Hg at the same temperature?
Sol. Given V1 = 1 litre P1 = 1 atm
V2 = ? P2 = 750 / 760 atm
P1V1 = P2V2
1 × 1 = 750 / 760 × V2
V2 = 1.0133 litre = 1013.3 ml
Prob 2. How large a balloon could you fill with 4g of He gas at 22°C and 720 mm of Hg?
Sol. Given, P = 720 / 760 atm, T = 295K, w = 4g
and m = 4 for He
PV = w / M RT
= 720 / 760 × V = 4/4 × 0.0821 × 295
∴ V = 25.565 litre
Prob 3. Calculate the temperature at which 28g N2 occupies a volume of 10 litre at
Sol. w = 28g, P = 2.46 atm, V = 10 litre, m = 28
Now, PV = w / M RT (R = 0.0821 litre atm K–1 mole–1)
T = 299.6 K
Prob 4. A gas occupies 300 ml at 27°C and 730 mm pressure what would be its volume at STP
Sol. V2 = 300 / 1000 litre, P2 = 730 / 760 atm, T2 = 300K
At STP, V1= ? P1= 1 atm, T1= 273K
P2V2 / T2 = P1V1 / T1 or V1 = 0.2622 litre
Volume at STP = 262.2 ml
Prob 5. In Victor Meyer’s experiment, 0.23g of a volatile solute displaced air which measures 112 ml at NTP. Calculate the vapour density and molecular weight of the substance.
Sol. Volume occupied by solute at NTP = Volume of air displaced at NTP
= 112 ml
For volatile solute PV = w / M RT
at NTP, P = 1 atm, T = 273 K
1 × 112/ 1000 = 0.23 / m × 0.0821 × 273
m = 46.02 and V.D. = 23.01
IIT Level Questions
Prob 6. A cylindrical balloon of 21 cm diameter is to be filled up with H2 at NTP from a cylinder containing the gas at 20 atm at 27°C. The cylinder can hold 2.82 litre of water at NTP. Calculate the number of balloons that can be filled up.
Sol. Volume of 1 balloon which has to be filled = 4/3 π (21/2)3 = 4851 ml
= 4.851 litre
Let n balloons be filled, then volume of H2 occupied by balloons = 4.851 × n
Also, cylinder will not be empty and it will occupy volume of H2 = 2.82 litre.
∴ Total volume occupied by H2 at NTP = 4.851 × n + 2.82 litre
∴ At STP
P2 = 1 atm Available H2
V1= 4.851 × n + 2.82 P2 = 20 atm
T1 = 273 K T2 = 300K
P1V1 / T2 = P2V2 / T2 V2 = 2.82 litre
or 1 × (4.85 1n + 2.82 / 273) = 20 × 2.82 / 300 ∴ n = 10
Prob 7. A 20g chunk of dry ice is placed in an empty 0.75 litre wire bottle tightly closed what would be the final pressure in the bottle after all CO2 has been evaporated and temperature reaches to 25°C?
Sol. w = 20g dry CO2 which will evaporate to develop pressure
m = 44, V = 0.75 litre, P = ? T = 298K
PV = W / m RT
P × 0.75 = 20 / 44 5 0.0821 × 298
P = 14.228 atm
Pressure inside the bottle = P + atm pressure
= 14.828 + 1 = 15.828 atm
Prob 8. The pressure of the atmosphere is 2 × 10–6 mm at about 100 mile from the earth and temperature is – 180°C. How many moles are there in 1 ml gas at this attitude?
Sol. Given, P = 2×10–6 / 760 atm
T = – 180 + 273 = 93 K
V = 1 ml = 1 / 1000 litre
PV = nRT
2 × 10–6 / 760 = n × 0.0821 × 93
n = 3.45 × 10–13 mole
Prob 9. 50 litre of dry N2 is passed through 36g of H2O at 27°C. After passage of gas, there is a loss of 1.20g in water. Calculate vapour pressure of water.
Sol. The water vapours occupies the volume of N2 gas i.e. 50 litre
∴ For H2O vapour V = 50 litre, w = 1.20g, T = 300K, m = 18
PV = w / m RT or P × 50 = 1.2 / 18 × 0.0821 × 300
∴ P = 0.03284 atm
= 24.95 mm
Prob 10. A mixture of CO and CO2 is found to have a density of 1.50g/litre at 30°C and 730 mm. What is the composition of the mixture?
Sol. For a mixture of CO and CO2, d = 1.50 g/litre
P = 730 / 760 atm, T = 303K
PV = w / m RT; PV w / Vm RT
730 / 760 = 1.5 / m × 0.0821 × 303 = ∴ 38.85
i.e. molecular weight of mixture of CO and CO2 = 38.85
Let % of mole of CO be a in mixture then
Average molecular weight = a × 28 + (100 – a) 44 / 100
38.85 = 28a + 4400 – 44a / 100
a = 32.19
Mole % of CO = 32.19
Mole % of CO2 = 67.81
Prob 11. The average speed of an ideal gas molecule at 27°C is 0.3 m sec–1. Calculate average speed at 927°C.
Sol. uav= √8RT / πM
Given uav = 0.3 m sec–1 at 300K
u1= 0.3 = √8R × 300 / πM
at T = 273 + 927 = 1200K
u2 = √8R × 1200 / πM
u2 / 0.3 = √1200 / 300 u2 = 0.6 m sec–1
Prob 12. Pure O2 diffuses through an aperture in 224 seconds, whereas mixture of O2 and another gas containing 80% O2 diffuses from the same in 234 sec. What is the molecular weight of gas?
Sol. For gaseous mixture 80% O2, 20% gas
Average molecular weight Mm = 32 × 80 + 20 × m / 100
Now, for diffusion of gaseous mixture and pure O2
rO2 / rm = √Mm / MO2 or VO2 / EO2 × Em / Vm = √Mm / MO2
∴ 1 / 224 × 234 / 1 = √Mm / 32 ∴ Mm= 34.92 ∴ 32 × 80 + 20 × m / 100 = 34.92
∴ m = 46.6
Prob 13. Calculate the pressure exerted by 10–23 gas molecules, each of mass 10–22 g in a container of volume one litre. The rms speed is 105 cm sec–1.
Sol. Given, n = 1023 m = 10–22g V = 1 litre = 103cc
urms = 105cm sec–1
PV = 1/3 mnu2rms
P × 103 = 1/3 × 10–22 × 1023 × (105)2
P = 3.3 × 107dyne cm–2
Prob 14. Calculate the temperature at which CO2 has the same rms speed to that of O2 at STP.
Sol. rrms of O2 = √3RT / M at STP, urms of O2 = √3R × 273 / 32
For CO2 urms CO2 = √3Rt / 44
Given both are same; 3R × 273 / 32 = 2RT / 44
∴ T =375.38 K = 102.38°C