States of matter (gases) Class XI

By:
Arunesh Gupta
PGT (Chemistry)
KV BKP (AFS)
States of Matter
Arunesh Gupta PGT (Chemistry) KV BKP (AFS)

MATTER
The substance that contains mass and occupies
space is known as Matter.
Atoms and molecules constitute matter.
It has various physical and chemical properties.
Matters are classified into solid, liquid and gas.
The force of interaction present in them varies
which gives rise to its physical properties.
Gases have the least force of interaction, while
solids have the highest force of interaction.

# Solids retain their fixed shape and volume, as the constituent
particles are rigid and non-compressible.
# There are negligible inter-particle spaces, and solids lack
flowability.
# Gases and liquids are fluid in nature (flow).
# They assume the size and shape of the container they are
captured and the particles can freely move past each other
because of the presence of a large number of inter-particle spaces.
# Apart from this, there is another important state of matter
known as Bose-Einstein Condensate where gases are cooled down
to lower densities.

Inter particle forces in matter
The forces of repulsion and attraction between the interacting particles
are commonly referred to as Intermolecular Forces. These interacting
particles have dipole moments that are permanent in nature. The
attractive forces between molecules are known as the van der Waal
Forces, which varies with different types of intermolecular reactions.
Van der Waal forces are the collective of (a) dipole-dipole interaction,
(b) dipole-induced dipole interaction and (c) dispersion forces.
Ion-dipole and ion-induced dipole are not van der Waal forces.
Hydrogen bonding is the strongest force of attraction.

The Different Types of Intermolecular Forces are:
Dipole-Dipole Interaction:
Polar molecules exhibit dipole-dipole interactions. They have
permanent dipole moments. The positive pole and negative pole
are attracted to the molecule. For example, H−Cl(l)
In HCl molecule, Cl is more electronegative than hydrogen. So,
the chlorine atom acquires a partial negative charge (𝛿−) while the
hydrogen atom acquires a partial positive charge (𝛿+). Thus,
dipole-dipole interaction takes place among them.

Dipole Induced Dipole Interaction:
The interaction between the non-polar molecules when a polarized
dipole is brought near to it is known as dipole-induced dipole
interaction. Here, a dipole is induced by a polar molecule in an atom
or electron disarrangement in a nonpolar molecule.
For example, in the presence of polar molecules, noble gases get
polarized, I2 dissolved in water, Chloroform dissolved in benzene etc.

Dispersion Forces or London Forces:
The weakest intermolecular forces are known as Dispersion
forces or London forces. It is a temporary attractive force that
takes place when two adjacent electron clouds of bonded atoms
form a temporary dipole by occupying positions between atoms
or non polar molecules.
The electron cloud of the molecule gets distorted for the generation
of an instantaneous dipole. The momentary dipole is produced in the
molecule in which one part of the molecule is more negative than the
other part. The momentary dipole produced induces a dipole in the
other molecule. Thus, the force of attraction between the induced
momentarily dipole is known as London or dispersion forces.

Hydrogen bond:
In the hydrogen bond, the dipole-dipole interaction between a
hydrogen atom with electronegative F, N, and O atoms in the form F-H,
N-H or O-H bonds is known as hydrogen bonding.
Chlorine (Cl) can also participate in the hydrogen bonding along with F,
N, and O in some cases. There is a coulombic interaction between the
hydrogen atom of one molecule with the lone-pair electrons of another
electronegative atom of the reactive molecule. Example: water is
liquid, ice is solid, sulphutic acid is viscous etc  due to inter
molecular Hydrogen bonds,

Intermolecular Forces Versus Thermal Energy:
The force of interaction has a tendency to bring the molecules closer. That is, the
solid possesses the strongest intermolecular force of attraction, while the gases
possess the least intermolecular force of attraction. The decreasing order is,
Solid > Liquid > Gas
The thermal energy is possessed by the molecule. The kinetic energy helps in the
movement of particles. The gas possesses the highest thermal energy, while the
solid possesses the least thermal energy. The decreasing order is,
Gas > Liquid > Solid

The gas laws:
Boyle’s law: It gives the relation between pressure-volume.
The law states that “the pressure of a given mass of an ideal gas is
inversely proportional to its volume at a constant temperature”.
Temperature & mass of gas  constant

Q1.) A canister of CO2 has a volume of 5.6 L and k = 0.54. What is the
pressure of the CO2?
Answer: Using Boyle’s law:
PV = k
so, P(5.6L) = 0.54 or, P = 0.54/5.6
Hence, P = 0.01 atm
So the pressure of a 5.6 L canister of CO2 would be 0.01 atm.
Q2.) A balloon full of helium (He) has a pressure of 3.6 atm and k = 1.31.
What is the volume of the balloon?
Answer:
Once again, plugging these value into the equation PV = k, we have,
(3.6atm) V = 1.31 or, V = 1.31/3.6
V = 0.36 L
So the volume of the helium-filled balloon at 3.6 atm would be 0.36 L.

Q3.) A canister of hexane has a volume of 2.9 L and a pressure of 1.45 atm.
The volume is then changed to 0.65L. What is the new corresponding
pressure?
Answer:
We know that for the same sample of gas, the product of the pressure and
volume in one state is equal to the product of the pressure and volume in
another state: P1V1 = P2V2
Given: P1 = 1.45 atm, V1 = 2.9 L, V2 = 0.65 L, P2 = ?
So, (2.9L)(1.45atm) = (0.65L)P2
Or, 13.05 = (0.65)P2 or, 13.05/0.65 = P2 or, P2 = 20.08 atm
The hexane have a new P = 20.08 atm (Ans.)
Q4.) 352 mL of chlorine under a pressure of 680 mm Hg are placed in a
container under a pressure of 1210 mm Hg. What is the volume of the
container in litres?
Answer:
From Boyle’s law the equation is P1V1 = P2V2
Given: P1 = 680 mm, V1 = 352 mL, P2 = 1210 mm, V2 = ?
(680 mm Hg)(352mL) = (1210 mm Hg)V2 on calculation, V2 = 198 mL
The same container of chlorine has a new volume of 198 mL. (Ans)

Q5.) A company makes a balloon that is made to be inflated to a volume no
greater than 3.0 L. If the balloon is filled with 2.3 L of helium at 1.67 atm, and
rises in the atmosphere until the pressure is 0.71 atm, will the balloon burst?
Ans.
We have, V1 = 2.3, P1 = 1.67, and P2 = 0.71.
Putting these values into the Boyle’s law equation P1V1 = P2V2
We get, (2.3L)(1.67atm) = V2(0.71atm)
V2 = 5.41 L
So yes, the balloon would burst.
Limitations of Boyle’s Law:
This law is applicable at moderate temperature & pressure.
At very low temperatures where particles are moving slowly though,
intermolecular forces become pronounced enough to have a significant
effect on the behaviour of the gas and Boyle’s law is no longer accurate.

Charle’s law (Temperature-Volume relationship)
Charles & Gay Lussac showed that for a fixed mass of a gas at constant
pressure, volume of a gas increases or deceases by
𝟏
𝟐𝟕𝟑.𝟏𝟓
of the original
volume of gas at 0℃ 𝐩𝐞𝐫 𝐝𝐞𝐠𝐫𝐞𝐞 𝐫𝐢𝐬𝐞 𝐨𝐫 𝐟𝐚𝐥𝐥 𝐨𝐟 𝐭𝐞𝐦𝐩𝐞𝐫𝐚𝐭𝐮𝐫𝐞.
If volume of a gas at 0℃ and t℃ are V0 & Vt respectively, then
Vt = V0 +
𝟏
𝟐𝟕𝟑.𝟏𝟓
V0 = V0 𝟏 +
𝟏
𝟐𝟕𝟑.𝟏𝟓
= V0
𝟐𝟕𝟑.𝟏𝟓+𝟏
𝟐𝟕𝟑.𝟏𝟓
At this stage, a new scale of temperature (kelvin scale or absolute scale) is
defined as
x℃ = 𝐱 + 𝟐𝟕𝟑. 𝟏𝟓 𝐊 ≈ 𝐱 + 𝟐𝟕𝟑 𝐊 𝐒𝐨, 𝐰𝐞 𝐜𝐚𝐧 𝐬𝐚𝐲: 𝟎℃ = 𝟐𝟕𝟑. 𝟏𝟓 K
So, at - 273.15 ℃ = 𝟎 𝐊.
At 0 K volume of a gas is zero, i.e. gases change into a solid or liquid state.
All gaseous calculations are done using kelvin scale (or also called
Thermodynamic scale of temperature)
Using kelvin scale: Vt = V0
𝐓𝐭
𝐓𝟎
⇒
𝐕𝐭
𝐕𝟎
=
𝐓𝐭
𝐓𝟎
⇒
𝑽
𝑻
= constant = k or, V = k. T

Charle’s law states that pressure remaining constant, the volume of a fixed
mass of a gas is directly proportional to its absolute temperature (temperature
in kelvin scale) i.e. V ∝ 𝐓 (𝐤𝐞𝐥𝐯𝐢𝐧)

Numerical problems:
(1) A balloon is filled to a volume of 2.40 L at a temperature of 20 °C. The
balloon is then heated to a temperature of 70 °C. Assuming the pressure
remains constant throughout, find the new volume of the balloon.
Solution: Given: V1 = 2.40 L T1 = 20 °C = 293 K T2 = 70 °C = 343 K V2 =?
We know V1/T1 = V2/T2
V2 = V1 ×T2 / T1
V2 = 2.40 × 343 / 293
V2 = 2.80 L
(2) A sample of a gas has an initial volume of 30.8 L and an initial
temperature of −67°C. What must be the temperature of the gas for its
volume to be 21.0 L?
Solution: Given: V1 = 30.80 L T1 = -67 °C = 206 K T2 =?
V2 = 21.0 L
We know V1/T1 = V2/T2 Hence, T2 = V2×T1 / V1 or, T2 = 30.80 × 206 / 21.0
T2 = 302 K =29 °C

Gay Lussac’s Law (Pressure-Temperature relationship)
(given by Joseph Gay Lussac)
It states that at constant volume, pressure of a fixed amount of a gas varies
directly with the temperature.
Mathematically, p ∝ 𝐓 𝐤𝐞𝐥𝐯𝐢𝐧 [𝐕𝐨𝐥𝐮𝐦𝐞 & 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐠𝐚𝐬 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭]
⇒
𝒑
𝑻
= constant = k Hence,
𝐏𝟏
𝐓𝟏
=
𝐩𝟐
𝐓𝟐

Avogadro’s law: It states that at constant temperature and pressure, the
volume of all gases constitutes an equal number of molecules.
In other words, this implies that in unchanged conditions of temperature and
pressure the volume of any gas is directly proportional to the number of
molecules of that gas. Mathematically, V ∝ n
Here, n is the number of moles of the gas. Hence, V= k n
The number of molecules in a mole of any gas is known as the Avogadro’s constant
NA = 6.022 x 1023. The standard temperature and pressure (STP) for temperature, T =
273.15 K while for the pressure it equals 1 bar or 105 pascals.
At these Standard Temperature Pressure (STP) values, one mole of a gas is
supposed to have the same volume. Now, no. of moles of gas n = m/M
According to Avogadro’s equation: V= k (m/M) or, M=k(m/V)
m/V= d (density); where M = molar mass of gas of mass m gram.
Therefore M =k D.
This means that at a constant temperature and pressure conditions, the molar mass
(M) of every gas is directly proportional to its density(d).

Ideal Gas: It is a hypothetical gas which follows Boyle’s law,
Charle’s law & Avogadro’s law strictly at all conditions. There is no
intermolecular forces between the molecules of an ideal gas.
Ideal gas equation: It is derived from Boyle’s law, Charle’s law &
Avogadro’s law in a single equation.
(i) Boyle’s law: V ∝
𝟏
𝐏
(T & no. of moles, n constant)
(ii) Charle’s law: V ∝ 𝐓 (P & n constant)
(iii) Avogadro’s law: V ∝ n (P & T constant)
Combining the equations: V∝
𝐧 𝐓
𝐏
⇒ V = R.
𝐧 𝑻
𝑷
⇒ 𝐏𝐕 = 𝐧𝐑𝐓 This is Ideal
gas equation & R is called Universal gas constant.
Value of ‘R’ depends on V, P & T

Unit of R in S.I. system

Different Units of ‘R’
From Ideal gas equation: PV = nRT or, R=PV/nT
(1) At STP (standard temperature and pressure) (T = 273.15 K & P = 1 bar),
1 mole of gas occupies 22.711 L. i.e. n = 1 mol & V = 22.711 L
To evaluate R we put the values .
R=PV/nT=1 bar×22.711 L/(1 mol × 273.15 K) =0.083 145 L atm⋅K-1mol-1
(2) In SI units, then pressure is measured in pascals (Pa) and volume is
measured in cubic metres (m3).
R=PV/nT=1.01325×105lPa×22.414×10-3lm3 / (1 mol × 273.15 K)
R = 8.3145 Pa⋅m3K-1mol-1

Qu: At 45° C and 900mm of Hg pressure a gas occupies 500ml
volume. What will be its pressure at a height where the
temperature is 15º C and volume is 750mL?
Solution:
Here we use the equation of Combined Gas Law, p1V1/T1 = p2V2/T2
Therefore we have: p2 = p1V1 T2/T1 V2
Hence, p1 = 900 mmHg, T1 = 45 +273= 318 K, V1 = 500 mL, T2=
15+273 = 288 K and V2 = 750 mL
p2 = (900 mm Hg) × (500mL) × 288 / 318 K × 750 mL = 543.96 mm
Hg
Therefore the height will be = 543.96 mm

Remember:
(i) PM = dRT
(d = density of gas)
(ii) M =
𝐰𝐑𝐓
𝐏𝐕
(iii) M =
𝐝 𝐑𝐓
𝑷

Dalton’s law of Partial Pressures: (John Dalton, 1801)
It states that the total pressure exerted by the mixture of non-
reactive gases is equal to the sum of the partial pressures of
individual gases under same temperature & in a fixed volume.
In a mixture of gases, the pressure exerted by the individual gas is called
partial pressure of that gas.
In a gas mixture of gases A, B, C, …. At constant temperature & fixed volume,
Total pressure, Ptotal = pA + pB + pC +….
Gases are generally collected over water & are therefore moist. The pressure
exerted by saturated water vapour in atmosphere is called aqueous tension.
So, pressure of dry gas is calculated as
Pdry gas = Ptotal – Aqueous tension. Arunesh Gupta PGT (Chemistry) KV BKP (AFS)

Dalton’s law of Partial Pressures:

SOLVED NUMERICAL PROBLEMS:
Problem.1) Determine the volume of occupied by 2.34 grams of carbon
dioxide gas at STP.
Solution:
1) Rearrange PV = nRT to this: V = nRT / P
Substituting the given data:
V = [(2.34 g / 44.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (273.0 K)] / 1.00 atm
V = 1.19 L (to three significant figures)
Problem 2) A sample of argon gas at STP occupies 56.2 litres. Determine
the number of moles of argon and the mass of argon in the sample.
Solution:
We know, PV = nRT or, n = PV / RT
n = [(1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K)]
or, n = 2.50866 mol
Mass in g = n x molar mass = 2.50866 mol x 39.948 g/mol = 100 g (Ans.)
Problem 3) At what temperature will 0.654 moles of neon gas occupy
12.30 litres at 1.95 atmospheres?
Solution:
We have, PV = nRT or, T = PV / nR
T = [(1.95 atm) (12.30 L)] / [(0.654 mol) (0.08206 L atm mol¯1 K¯1)] T
= 447 K (Ans.)

Problem 4) A 12.0 g sample of gas occupies 19.2 L at STP. What is the
molecular weight of this gas?
Solution:
We know, PV = nRT: Putting the values,
(1.00 atm) (19.2 L) = (n) (0.08206) (273 K) or, n = 0.857 mol
12.0 g / 0.857 mol = 14.0 g/mol (Ans)
OR
Since it is at STP, we can also use molar volume:
(19.2 L / 12.0 g) = (22.414 L / x ) or, 19.2x = 268.968 or, x = 14.0 g/mol (Ans)
Problem 5) 5.600 g of solid CO2 is put in an empty sealed 4.00 L container at
a temperature of 300 K. When all the solid CO2 becomes gas, what will be
the pressure in the container?
Solution:
1) Determine moles of CO2:
5.600 g / 44.009 g/mol = 0.1272467 mol
2) Use PV = nRT
(P) (4.00 L) = (0.1272467 mol) (0.08206 L atm mol¯1 K¯1) (300 K)
P = 0.7831 atm (to four sig figs) (Ans)

Problem 6) A sample of nitrogen occupies a volume of 1.0 L at a pressure of
0.5 bar at 40°C. Calculate the pressure if the gas is compressed to 0.225 mL
at –6°C.
Solution:
P1 = 0.5 bar P2 = ?
V1 = 1.0 L V2 = 0.225 mL = 0.225 × 10–3 L T1 = 40+273 = 313 K T2 = – 6 + 273 =
267 K
According to combined gas law equation,
P1V1/T1 = P2V2/T2
P2 = (P1V1T2)/(T1V2) = (0.5 x 1x 267)/(313x 0.225 ×10–3) = 1895.6 bar (Ans)
Problem 7) At 25°C and 760 mm of Hg pressure, a gas occupies 600 mL
volume. What will be its pressure at a height where temperature is 10°C and
volume of the gas is 640 mL.
Solution:
P1=760 mm Hg P2 = ?
V1=600 mL V2 = 640 mL T1=273 + 25 = 298 K T2= 273 + 10 = 283 K
According to combined gas law equation,
P1V1/T1 = P2V2/T2
P2 = (P1V1T2)/(T1V2) = [(760 mm Hg) (600 mL) (283 K)] / [(298K)(640 mL) ]
= 676.6 mm Hg (Ans)

Problem 8) Calculate the moles of hydrogen (H2) present in a 500 mL sample
of hydrogen gas at a pressure of 1 bar and 27°C.
Solution:
1 atm = 1.01325 bar = 1.01325 × 105Pa = 101.325 kPa & or 1 bar = 0.987 atm
According to ideal gas equation,
PV = nRT
P = 1 bar = 1 atm = 101 x 103 Pa V = 500 mL = 500 cm3 = 500 × 10–6 m3
T = 27 + 273 = 300 K, R = 8.314 J mol–1K–1
Now, n = PV/(RT) = 101 x 103 x 500 × 10–6 / ( 8.314 x 300)
= 0.02 moles (Ans)
Problem 9) Calculate the volume occupied by 4.045 × 1023 molecules of
oxygen at 27°C and having a pressure of 0.933 bar.
Solution:
Here, number of molecules = 4.045 × 1023
P = 0.933 bar = 0.933 atm = 0.933 x 101 x 103 Pa, T = 27 + 273 = 300 K
R =8.314 J mol–1K–1
We will use PV = nRT. or V = nRT/P
Let us first calculate the number of moles, n.
We know that number of moles, n = No. of molecules / 6.022 x 1023
= 4.045 × 1023/ 6.022 x 1023 = 0.672 mol
V = nRT/P = 0.672 x 8.314 x 300 / ( 0.933 x 101 x 103)
= 0.0177 m3 (Ans)

Problem 10) A discharge tube of 2 L capacity containing hydrogen gas was
evacuated till the pressure inside is 1 × 10–5atm. If the tube is maintained at a
temperature of 27°C, calculate the number of hydrogen molecules still present
in the tube.
Solution:
Step I. To calculate the number of moles of hydrogen.
Here P = 1 × 10–5 atm = 1 × 10–5 x 1.01 x 105 Pa = 101 x 10-2 Pa
V = 2 litres =2 000 cm3 = 2000 x 10-6 m3 = 2 x 10-3 m3 & R = 8.314 J mol–1K–1
T = 27 + 273 = 300 K n = ?
According to the gas equation, PV = nRT
or n = PV/RT = 101 x 10-2 x 2 x 10-3/ (8.314 x 300) = 8.1 x 10-7 mole
Step II. To calculate the number of hydrogen molecules.
We know that 1 mol of hydrogen = 6.022 × 1023 molecules
8.1 x 10-7 mole = 6.022 × 1023 x 8.1 x 10-7 = 48.77 x 1016 molecules (Ans)
Problem 11) Calculate the mass of 120 mL of N2 at 150°C and 1 × 105 Pa
pressure.
Solution:
According to ideal gas equation, PV = nRT =(m/M)RT ⇒ m = PVM/(RT)
P = 105 Pa , V = 120 ×10–6 m3 M = 28,
T = 273 + 150 = 423 K & R = 8.314 Nm K–1 mol–1
Mass of gas m = PVM/(RT) =(105 x 120 × 10–6 x 28)/(8.314 x 423)
= 0.0955 g (Ans)

Postulates of K.T. of gases (contd.)
(6) The collision between the gas molecules are perfectly elastic.
(7) There is no effect of gravity on the gas molecules.
From Kinetic theory of gases, mathematical expression called kinetic
gas equation is derived as PV =
𝟏
𝟑
mN𝒖𝟐
where m = mass of one gas molecule
N = total no. of gas molecules,
u = root mean square speed of gas molecules
Mean K.E. of a molecule =
1
2
m u2
(For 1 mole of a gas molecules = N = NA = 6.022 x 1023 molecules)
Hence, K.E. of 1 mole of a gas = ½ mNu2
PV =
2
3
(½ mNu2) =
2
3
K.E.
For 1 mole of an ideal gas: PV = RT
Hence, PV =
2
3
K.E. = RT or K.E. =
𝟑
𝟐
RT ,
We can say K.E. ∝ 𝐓 (absolute scale)

KINETIC THEORY of GASES:
Also, K.E. per molecule =
𝟑
𝟐
𝐑𝐓
𝐍
=
𝟑
𝟐
kT
where k =
𝐑
𝐍
= Boltzmann constant
# K.E. = ½ mu2 or, K.E. ∝ u2 and K.E. ∝ T
So, u2 ∝ 𝐓 or, u ∝ 𝐓
Hence, molecular speed of a gas is directly proportional to
the square root of absolute temperature.

(1) The most probable speed (vmp) is the one which informs about the
speed possessed by the maximum number of molecules of the gas.
The most probable speed is the maximum value on Maxwell's distribution
plot.
So, vmp ∝ 𝑇
On increasing temperature most probable speed of gas molecules increase.
Also, vmp ∝
1
𝑀
We can say, N2 (molar mass = 28) has more most probable speed than that of
Cl2 (molar mass = 71)

(2) The average speed (vav) is the sum of the speeds of all the molecules
divided by the number of 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, in which
n1 molecules have a speed of v1, n2 molecules have a speed of v2, , …..…
nn molecules have a speed of vn.
The average speed or vav = average of all such speed terms.
The average or mean speed = vav=
𝒗𝟏+𝒗𝟐+𝒗𝟑……………..𝒗𝒏
𝑵
Here, N = n1 + n2 + n3 + ….....+nn
And vav =
𝟖 𝑹𝑻
𝝅 𝑴
where M = molar mass of the gas. R = 8.314 J/mol/K

(3) The root-mean-square speed (vrms) is the measure of the
speed of particles in a gas, defined as the square root of the average
velocity-squared of the molecules in a gas.
vrms =
𝑣1
2 +𝑣2
2 +𝑣2
+⋯..
𝑁
1
2

Graphically, is shown by Maxwell – Boltzmann distribution curves. The most
probable speed is the maximum value on Maxwell's distribution plot.
The root mean square speed, average speed & the most probable speed have
following relationship.
vrms > vav > vmp
The ratio between the above three speeds is vmp : vav : vrms = 1 : 1.128 : 1.224

REAL GAS:
A real gas is defined as a gas that at all standard pressure and temperature conditions
does not obey gas laws. Real gases can be defined as non ideal gases whose molecules
occupy a given amount of space and have the ability to interact with each other.
(1) Why do gases deviate from the ideal gas behavior?
(2) What are the conditions under which gases deviate from ideality?
Two assumptions of kinetic theory of gases do not hold good.
(i) There is no force of attraction between the gas molecules.
(ii) Volume of the molecules of a gas is negligibly small in comparison to the
space occupied by the gas.

If assumption (i) is correct, we can not explain why gases liquefy
when cooled & compressed. Gases are compressible but liquids are
not. Liquids occupy less volume compared to the gases.
If assumption (ii) is correct, the pressure vs volume graph and
experimental data (for real gas) & that of theoretically calculated
from Boyle’s law (ideal gas equation) should coincide.
So, real gases show deviations from ideal gas law because molecules interact
with each other.
Therefore , for a real gas, ideal pressure (Pi) & ideal volume (Vi) is required to
correct from ideal gas behavior as Pi.Vi = nRT.
Pi & Vi are ideal pressure & ideal volume of a gas.
Thus, after pressure correction & volume correction, van der Waals equation is
written for a real gas as
(P + a
𝐧𝟐
𝐕𝟐) (V-nb) = nRT (for n moles of a real gas) &
(P +
𝐚
𝐕𝟐) (V – b) = RT (for 1 mole of a real gas)

Volume correction (vcorr):
The actual free space available inside the vessel
for the movement of molecules is not the
volume of the container but Vi is the actual
volume of the container (V) minus effective
volume (b)
(b is called as co-volume or excluded volume)
It is a van der Waals constant)
Volume correction: vcorr ∝ 𝒏
Or, v = n.b ( for n moles of a gas)
Hence, Vi = (V – vcorr)
So, Vi = (V – b) (for 1 mole of a gas)
or, Vi = (V – nb) (for n moles of a gas)

Pressure correction:
All other gas molecules on all sides attract a molecule in the interior of
a gas. These attractive factors cancel each other out. However, a
molecule strikes the vessel’s wall is only drawn by molecules on one
side. As a result, it feels compelled to turn inside.
As a result, it hits the wall at less velocity, and the real gas pressure P,
is lower than the ideal pressure (Pi).
So, the actual pressure P is less than Pi by a quantity pcorr (corrected)
We have, P = Pi–pcorr or, Pi = P + pcorr

Pressure correction (pcorr) :
We have Pi = P + pcorr
We know, density of gas (d) =
mass of gas (m)
volume of gas (V)
or, d ∝
mass of gas
molar mass of gas
Volume of gas (V)
or, d ∝
no.of moles of gas (n)
Volume of gas (V)
⇒ 𝒅 ∝
𝐧
𝐕
The gas molecule is equally attracted by other molecules surrounding it in the
bulk and has inward attraction at the inner surface of the vessel during collision.
So, 𝑝𝑐𝑜𝑟𝑟 ∝ 𝑑 (at the bulk)
Also, pcorr ∝ 𝑑 (at the inner walls of vessel)
⇒ pcorr ∝ 𝑑2
⇒ pcorr ∝
𝑛
𝑉
2
⇒ pcorr = a
n
V
2
We conclude Pi = P + a
n
V
2
or Pi = P + a
𝐧𝟐
𝐕𝟐

So, considering the pressure correction & volume correction for a real
gas, the ideal gas equation is rewritten as van der Waals equation:
Pi = P + a
𝒏𝟐
𝑽𝟐 and Vi = (V – nb) & Pi Vi = nRT
Here, n is the no. of moles of a real gas.
‘a’ and ‘b’ are van der Waals constants or van der Waals
parameters. Their values depends on the characteristic of a gas.

Units of ‘a’ & ‘b’
‘a’ ⇒ pcorr = a
𝐧𝟐
𝐕𝟐 ⇒ a =
𝐩𝐜𝐨𝐫𝐫𝐕𝟐
𝐧𝟐 =
𝐚𝐭𝐦. 𝐋𝟐
𝐦𝐨𝐥𝟐 = atm L2 mol-2
Or unit of ‘a’ is bar dm6 mol-2 or Pa m6 mol-2 (SI unit)
‘b’ ⇒ vcorr = n b ⇒ b =
𝐯𝐜𝐨𝐫𝐫
𝐦𝐨𝐥
⇒ Unit of ‘b’ = L mol-1 or dm3 mol-1 or m3 mol-1 (SI unit)
Significance of van der Waals constants: ‘a’ and ‘b’:
‘a’ ⇒ it is a measure of magnitude of attractive forces among the molecules of
the gas. It is independent of temperature & pressure.
Greater is the value of ‘a’ ⇒ larger is the intermolecular force of
attraction & the gas is easily liquefied.
‘b’ ⇒ It is the measure of the effective size of the gas molecules.
Its value is four times the actual volume of the gas molecule,

A real gas shows ideal gas behavior when pressure approaches zero &
intermolecular forces are practically negligible.
It is also in case when temperature is very high.
Under these conditions volume corrections & pressure corrections are
tends to zero & van der Waals equation obeys ideal gas equation.
The deviation from ideal gas behavior can be measured by compressibility
factor Z, which is the ratio of PV & nRT.
Mathematically, Z =
𝐏𝐕
𝐧𝐑𝐓
(i) If Z = 1, PV = nRT. It is an ideal gas. Example: N2 has Z = 1 at 50oC at a
pressure 1 to 100 atm.
The temperature at which a real gas behaves like an ideal gas at an
appreciable range of pressure is called Boyle’s temperature or Boyle point (TB)
TB =
𝒂
𝒃𝑹
(ii) When Z > 1, the gas shows positive deviation from ideal gas behavior & is
less compressible. i.e. PV > nRT Example: H2, He etc.
(iii) When Z < 1, the gas shows negative deviation from ideal gas behavior. i.e.
PV < nRT & is more compressible & easily liquefied. Example: CO2, HCl, NH3
etc.

We can further derive:
Z =
𝑷 𝑽𝒓𝒆𝒂𝒍
𝑹𝑻
⇒ P Vreal = Z nRT -------- (i)
& for ideal gas: P Videal = nRT ------ (ii) at same temperature &
pressure
Dividing (i) by (ii) we get,
𝑽𝒓𝒆𝒂𝒍
𝑽𝒊𝒅𝒆𝒂𝒍
= Z
So, compressibility factor ’Z’ is the ration of actual
molar volume of a gas to the ideal molar volume of it, if
it were an ideal gas at that temperature & pressure.

THANK YOU

States of matter (gases) Class XI

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