NCERT Solutions for Class 11 Chemistry Chapter 5: States of Matter

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5.1. What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30°C?

Ans. Given,

Initial volume of a gas (V₁) = 500 dm³

Initial pressure of a gas (P₁) = 1 bar at 30 °C.

Final volume of a gas (V₂) = 200 dm³

Final pressure of a gas (P₂) = ? at 30 °C.

Since the temperature is kept constant, the final pressure (P₂) can be calculated using Boyle’s Law.

Thus,
According to Boyle’s law,

P₁V₁ = P₂V₂

⇒ 1 bar × 500 dm³ = P₂ × 200 dm³

$$\Rarr\space\frac{1\space\text{bar}× 500\space\text{dm}^{3}}{200\space\text{dm}^{3}}=\text{P}_2$$

P₂ = 2.5 bar

Hence, the minimum pressure required is 2.5 bar.

5.2. A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?

Ans. Given,
Initial volume of a gas (V₁) = 120 mL

Initial pressure of a gas (P₁) = 1.2 bar at 35 °C.

Final volume of a gas (V₂) = 180 mL

Final pressure of a gas (P₂) = ? at 35 °C.

Since the temperature is kept constant, the final pressure (P₂) can be calculated using Boyle’s law.

Thus,

According to Boyle’s law,

P₁V₁ = P₂V₂

$$\Rarr\space 1.12\text{bar}×120\space\text{mL}=\text{P}_2× 180 \text{mL}\\\Rarr\frac{1.2 \text{bar}×120\space\text{mL}}{180\space\text{mL}}=\text{P}_2$$

P₂ = 0.8 bar.

Hence, the final pressure of the gas would be 0.8 bar.

5.3. Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressure p.

Ans. The equation of state is given by,

pV = nRT ...(i)

Where,

p = Pressure of gas

V = Volume of gas

n = Number of moles of gas

R = Gas constant

T = Temperature of gas

From equation (i) we have,

$$\frac{n}{\text{V}}=\frac{\text{P}}{\text{RT}}\\\text{Ideal gas equation in terms of density:}\\\text{n}=\frac{m}{\text{M}}\\\text{Replacing n with}\frac{m}{n}\text{we have}\\\frac{m}{\text{MV}}=\frac{p}{\text{RT}}\space\text{...(ii)}$$

Where,

m = Mass of gas

M = Molar mass of gas

$$\text{But}\space\frac{m}{v}=d\space\text{(d = density of gas)}\\\text{Thus, from equation (ii), we have}\\\frac{d}{\text{M}}=\frac{p}{\text{RT}}\\\Rarr\space\text{d}=\bigg(\frac{\text{M}}{\text{RT}}\bigg)p$$

Molar mass (M) of a gas is always constant and therefore, at constant temperature (T),

$$\frac{\text{M}}{\text{RT}}\space\text{= constant.}$$

d = (constant) p

⇒ d ∝ p

Hence, at a given temperature, the density (d) of gas is proportional to its pressure (p).

5.4. At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

Ans. Density (d) of the substance at temperature (T) can be given by the expression,

$$d=\frac{\text{Mp}}{\text{RT}}$$

Now, density of oxide (d1) is given by,

$$d_1=\frac{\text{M}_1\text{P}_1}{\text{RT}}$$

Where, M₁ and p₁ are the mass and pressure of the oxide respectively.

Density of dinitrogen gas (d₂) is given by,

$$d_2=\frac{\text{M}_2\text{P}_2}{\text{RT}}$$

Where, M₂ and p₂ are the mass and pressure of the oxide respectively.

According to the question, it is given:

p₁ = 2 bar

p₂ = 5 bar

Molecular mass of nitrogen, M₂ = 28 g/mol

Molecular mass of oxide, M₁ = ?

Since,

d₁ = d₂

∴ M₁p₁ = M₂p₂

$$\text{Now}\space\text{M}_1=\frac{M_2P_2}{P_1}\\=\frac{(28×5)}{2}\\$$

= 70 g/mol

Hence, the molecular mass of the oxide is 70 g/mol.

5.5. Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

Ans. Given :

m_A = 1 g

p_A = 2 bar

m_B = 2g

p_B = (3 – 2) = 1 bar

(Since total pressure is 3 bar)
For ideal gas A, the ideal gas equation is given by,

p_AV = n_ART ...(i)

Where p_A and n_A represent the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is given by
p_BV = n_BRT ...(ii)

Where, p_B and n_B represent the pressure and number of moles of gas B.

[V and T are constants for gases A and B]

From equation (i), we have

$$P_{AV}=\frac{m_A}{M_A}\text{RT}\\\Rarr\space\frac{\text{P}_A \text{M}_A}{m_\text{A}}=\frac{\text{RT}}{\text{V}}\space\text{...(iii)}$$

From equation (ii), we have

$$\text{p}_\text{B}=\frac{m_B}{M_B}\text{RT}\\\Rarr\space\frac{P_BM_B}{m_B}=\frac{RT}{V}\space\text{...(iv)}$$

Substituting the given values in equation (v), we get

$$\frac{2×\text{M}_\text{A}}{1}=\frac{1×\text{M}_\text{B}}{2}$$

⇒ 4M_A = M_B

Thus, a relationship between the molecular masses of A and B is given by,

4M_A = M_B

5.6. The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminium reacts?

Ans. The given chemical reaction is,

2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

2 moles of Aluminium produce 3 moles of dihydrogen

At STP (273.15 K and 1 atm). 54 g (2 × 27 g) of aluminium produce 3 × 22400 mL of dihydrogen.

[1 mole = molar mass = 22.4 L of volume]

$$\therefore\space\text{0.15 g Al gives}\frac{3×22400×0.15}{54}\text{mL}$$

186.67 mL of H₂ = 186.7 mL

S.T.P. condition V₁ = 186.7 mL, V₂ = ?

p₁ = 1.013 bar, p₂ = 1 bar

T₁ = 0 + 273 = 273 K, T₂ = 20 + 273 = 293 K

According to gas equation

$$\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}\\\text{or}\space\text{V}_2=\frac{p_1V_1T_2}{p_2T_1}\\\text{V}_2=\frac{1.013\text{bar}×186.7\space\text{mL}×293\text{K}}{1 \text{bar}×273\space\text{k}}$$

= 203 mL

Hence, 203 mL of dihydrogen will be released.

5.7. What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27 °C ?

Ans. Given, V = 9 dm³, T = 27 + 273 = 300 K

R = 8.314 × 10^–2 dm³ bar K^–1 mol^–1

It is known that,

$$\text{Number of moles =}\frac{\text{Given mass}}{\text{Molar Mass}}$$

Therefore,

Molar mass of methane,

CH₄ = 12 + 4 = 16 g mol^–1

Number of moles of methane

$$=\frac{\text{3.2\space\text{g}}}{16\space\text{g}\space\text{mol}^{-1}}=0.2\space\text{mol}$$

Now, the partial pressure exerted by 3.2 g of methane can be calculated using ideal gas equation as,

PV = nRT

$$p\text{CH}_4=\frac{n\text{RT}}{\text{V}}\\=\frac{(0.2\text{mol})(8.314×10^{\normalsize-2}\text{dm}^{3}\text{bar}\space \text{K}^{-1}\text{mol}^{-1})(300K)}{(9\space\text{dm}^{3})}$$

= 0.5543 bar.

Similarly,

Molar mass of carbon dioxide,

CO₂ = (12 + 2 × 16) = 44 g mol^–1

Number of moles of carbon dioxide

$$=\frac{4.4\space\text{g}}{44 \space\text{g}\space\text{mol}^{\normalsize-1}}=0.1\text{mol}$$

Now, the partial pressure exerted by 4.4 g of carbon dioxide can be calculated using ideal gas equation as,

$$p\text{co}_2=\frac{n\text{RT}}{\text{V}}\\=\frac{(0.1\space\text{mol})(8.314×10^{\normalsize-2}\space\text{dm}^{3}\text{bar K}^{-1}\text{mol}^{\normalsize-1})(300\text{K})}{(9\space\text{dm}^{3})}$$

= 0.2771 bar

Total pressure exerted by the mixture

p=PCH₄+PCO₂

= (0.5543 + 0.2771) bar = 0.8314 bar

= 0.8314 × 10⁵ Pa = 8.314 × 10⁴ Pa.

Hence, the pressure exerted will be 8.314 × 10⁴ Pa.

5.8. What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C?

Ans. Let the partial pressure of H₂ and O₂ be PH₂ and PO₂, respectively.

The partial pressure of H₂ in 1 L vessel,

Given P₁ = 0.8 bar, V₁ = 0.5 L, P₂ = ? and V₂= 1 L

According to Boyle’s law,

P₁V₁ = P₂V₂

⇒ 0.8 bar × 0.5 L = P₂ × 1 L

Hence, P₂= 0.40 bar, i.e., PH₂ = 0.40 bar

The partial pressure of O₂ in 1 L vessel,

Given P₁′ = 0.7 bar, V₁′ = 2 L

P₂′ = ? and V₂′ = 1 L

According to Boyle’s law,

P′₁V′₁= P′₂V′₂

⇒ 0.7 bar × 2 L = P₂ × 1 L

Hence, P′₂ = 1.4 bar, i.e., P_O₂

= 1.4 bar

Thus, total pressure will be,

P_total = PH₂+ PO₂

= [0.40 + 1.40] bar

= 1.80 bar

Thus, the total pressure of the gaseous mixture in the vessel will be 1.80 bar.

5.9. Density of a gas is found to be 5.46 g/dm³at 27 °C at 2 bar pressure. What will be its density at STP?

Ans. Given,

d₁ = 5.46 g/dm³

p₁ = 2 bar

T₁ = 27°C = (27 + 273) K = 300 K

p₂ = 1 bar

T₂ = 273 K

d₂= ?

The density (d₂) of the gas at STP can be calculated using the equation,

$$d=\frac{\text{M}p}{\text{RT}}\\\therefore\frac{d_1}{d_2}=\frac{\frac{\text{Mp}_1}{\text{RT}_1}}{\frac{\text{Mp}_2}{\text{RT}_2}}\\\text{For same gas, M and R are constant, therefore,}\\\Rarr\space\frac{d_1}{d_2}=\frac{p_1T_2}{p_2T_1}\\\Rarr\space d_2=\frac{p_2T_1d_1}{p_1T_2}\\=\frac{1×300×5.46}{2×273}$$

= 3 g dm^–3

Hence, the densityof the gas at STP will be 3 g dm^–3.

5.10. 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

Ans. Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × 10^–3 L

R = 8.314 × 10^–2 bar L K^–1 mol^–1

T = 546 °C = 546 + 273 = 819 K

m = 0.0625 g

Let the molar mass be M.

∴ Number of moles of phosphorus vapours

$$n=\frac{m}{\text{M}}=\frac{0.0625}{\text{M}}\text{mol}\\\text{From ideal gas equation}\\\text{pV}=n\text{RT}=\frac{\text{mRT}}{\text{M}}\\\text{or}\space\text{M}=\frac{\text{mRT}}{\text{pV}}\\=\frac{(0.0625g)(8.314×10^{\normalsize-2 }\text{bar}\text{L}\text{K}^{\normalsize-1}\text{mol}^{\normalsize-1})(819\text{K})}{(0.1\space\text{bar})(34.05×10^{\normalsize-3}\text{L})}$$

= 1249.8 g mol^–1

5.11. A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out?

Ans. Let us assume that volume of the vessel = V cm³
Then, the volume of air in the flask at 27°C =
V cm³
Now,
V₁ = V
V₂ = ?
Now, Initially, the temperature of the air is 27°C = 300 K, when the air has been expelled the temperature of the flask is 477°C = 750 K. Thus,
T₁ = 300 K and T₂ = 750 K
According to Charles law,

$$\frac{\text{V}_1}{\text{T}_1}=\frac{\text{V}_2}{\text{T}_2}\\\Rarr\space \text{V}_2=\frac{\text{V}_1\text{T}_2}{\text{T}_1}\\=\frac{750\text{V}}{300}$$

Therefore, volume of air expelled out

= 2.5 V – V = 1.5 V

$$\text{Hence, fraction of air expelled out =}\frac{1.5\text{V}}{2.5\text{V}}=\frac{3}{5}$$

5.12. Calculate the temperature of 4.0 mol of a gas occupying 5 dm³ at 3.32 bar. (R = 0.083 bar dm³ K^–1 mol^–1).

Ans. Given,

n = 4 mol, V = 5 dm³, P = 3.32 bar, R = 0.083 bar dm³ K^–1 mol^–1

Using ideal gas equation,

PV = nRT

3.32 bar × 5 dm³ = 4 × 0.083 bar dm³ K^–1 mol^–1× T

$$\text{T}=\frac{3.32\text{bar}×5\space\text{dm}^{3}}{4×0.083\text{bar\space dm}^{3}\space\text{K}^{\normalsize-1}\text{mol}^{\normalsize-1}}$$

T = 50 K

Hence, the temperature of the gas is 50K.

5.13. Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

Ans. Molar mass of dinitrogen (N₂) = 2 × 14 = 28 g

According to mole concept,

The number of molecules of N₂ in 28 g

= 6.023 × 10²³ molecules
The number of molecules of N2 in 1.4 g

$$=\frac{6.023×10^{23}×1.4}{28}$$

= 3.011 × 10²² molecules

Atomic number of nitrogen (N) = 7

Since, one atom of nitrogen contains 7 electrons, one molecule of nitrogen (N₂)contains = 2 × 7 = 14electrons

Number of electrons in 1 molecule of N₂

= 7 × 2 = 14

Number of electrons in 3.01 × 10²² molecules of N₂

= (14 × 3.01 × 10²²)

= 4.214 × 10²³

Hence, the total number of electrons present are 4.214 × 10²³.

5.14. How much time would it take to distribute one Avogadro number of wheat grains, if 10¹⁰ grains are distributed each second ?

Ans. One Avogadro’s number = 6.023 × 10²³ particles

One Avogadro number of wheat grains

= 6.023 ×10²³grains

According to the question,

Time required to distribute 1010 grains = 1 s

Time required to distribute 6.02 × 10²³ grains

$$=\frac{6.023×10^{23}}{10^{10}}\text{s}\\=6.022×10^{13}\text{s}\\=\frac{6.022×10^{13}}{60×60×24×365}\text{years}$$

= 19.096 × 10⁵ years or 1.909 × 10⁶ years

Hence, time taken would be 1.909 × 10⁶ years.

5.15. Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm³ K^–1 mol^–1.

Ans. Given,

V = 1 dm³, T = 27 °C = 300 K, R = 0.083 bar dm³ K^–1mol^–1

Mass of dioxygen (O₂) = 8 g

Thus, Number of moles of O₂ (nO₂)

$$=\frac{8\text{g}}{32\text{g}/\text{mol}}=0.25\space\text{mol}$$

Thus, number of moles of H₂ (n_H2)

$$=\frac{4\text{g}}{2\text{g}/\text{mol}}=2\space\text{mol}$$

Mass of dihydrogen (H₂) = 4 g

Therefore, total number of moles in the mixture

= n_O2+n_H2

= 2 mol + 0.25 mol

= 2.25 moles

According to ideal gas equation,

PV = nRT

$$P=\frac{n\text{RT}}{\text{V}}\\\text{P}=\frac{2.25\space\text{mol}× 0.083\space\text{bar}\space\text{dm}^{3}\text{K}^{\normalsize-1}\text{mol}^{\normalsize-1}×300\text{K}}{1\space\text{dm}^{3}}$$

P = 56.025 bar

Hence, the total pressure of the mixture is 56.025 bar.

5.16. Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m^–3 and R = 0.083 bar dm3 K^–1 mol^–1).

Ans. Given, Radius of the balloon = 10 m

Density of air = 1.2 kg m^–3

Since the balloon is spherical in nature.

$$\text{Hence, the volume of the balloon =}\frac{4}{3}\neq \text{r}^{3}\\=\frac{4}{3}×\frac{22}{7}×(10\space\text{m})^{3}$$

= 4190.5 m³

The mass of displaced air

= 4190.5 m³ × 1.2 kg m^–3

= 5028.6 kg

Now, mass of helium (m) inside the balloon is given by,

$$\text{m}=\frac{\text{M}p\text{V}}{\text{RT}}$$

Here,

M = 4 × 10^–3kg mol^–1

p = 1.66 bar

V = Volume of the balloon
= 4190.5 m³

R = 0.083 bar dm³K^–1mol^–1

T = 27°C = 300 K

$$\text{Then \space m =}\frac{4×10^{\normalsize-3}×1.66×4190.5×10^{3}}{0.083×300}$$

= 1117.5 kg (approx)

Now, total mass of the balloon filled with helium

= (100 + 1117.5) kg

= 1217.5 kg

Hence, Pay load = (5028.6 – 1217.5) kg

= 3811.1 kg

Hence, the pay load of the balloon is 3811.1 kg.

5.17. Calculate the volume occupied by 8.8 g of CO₂ at 31.1°C and 1 bar pressure. R = 0.083 bar L K^–1 mol^–1.

Ans. Given,

ω = 8.8 g, R = 0.083 bar LK^–1mol^–1, T = 31.1°C

= 31.1 + 273 K, P = 1 bar

Using the Ideal gas equation, PV = nRT

$$\text{PV}=\frac{\omega\text{RT}}{\text{M}}\\\text{or}\space\text{V}=\frac{\omega\text{RT}}{\text{MP}}\\\text{V}=\frac{8.8\text{g}}{44\text{g}/\text{mol}}×\frac{0.083\space\text{bar}\text{LK}^{\normalsize-1}\text{mol}^{\normalsize-1}×(273+31.1)\text{K}}{\text{1 \text{bar}}}$$

= 5.05 L

Hence, the volume occupied by CO₂ gas is 5.05 L.

5.18. 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?

Ans. Given,

For H₂ gas,	For unknown gas,
p₁= p	p₂ = p
V₁ = V	V₂ = V
T₁ = 17+273 = 290 K	T₂ = 95 + 273 = 368 K
m₁=0.184 g	m₂ = 2.9
M₁ = 2 g mol^–1	M₂ = ?

From ideal gas equation,

Volume (V) occupied by dihydrogen is given by

$$\text{V}=\frac{m}{\text{M}}\frac{RT}{p}\\=\frac{0.184}{2}×\frac{\text{R}×290}{p}$$

Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown gas can be calculated as :

$$\text{V}=\frac{m}{\text{M}}\frac{\text{RT}}{p}\\=\frac{2.9}{\text{M}}×\frac{\text{R}×368}{p}$$

According to the question,

$$\frac{0.184}{2}×\frac{R×290}{p}=\frac{2.9}{\text{M}}×\frac{\text{R}×368}{p}\\\Rarr\space\frac{0.184×290}{2}=\frac{2.9×368}{\text{M}}\\\Rarr\text{M}=\frac{2.9×368×2}{0.184×290}\\=40\space g\space\text{mol}^{\normalsize-1}$$

Hence, the molar mass of unknown gas is 40 gm mol^–1.

5.19. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Ans. Given,

Total pressure of the mixture, p_total = 1 bar

Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g.

Then, the number of moles of dihydrogen,

$$n_{H_{2}}=\frac{20}{2}=10\space\text{and the number of moles of dioxygen,}\space n_{O_{2}}=\frac{80}{32}=2.5\space\text{moles}$$

Then, partial pressure of dihydrogen,

$$\text{P}_{\text{H}_2}=\frac{n_{H_{2}}}{n_{H_{2}}+n_{O_{2}}}×P_{total}\\=\frac{10}{10+2.5}×1$$

= 0.8 bar

Hence, the partial pressure of dihydrogen is 0.8 bar.

5.20. What would be the SI unit for the quantity _pV²T²/n ?

Ans. The SI unit for pressure, p is Nm^–2.

The SI unit for volume, V is m³.

The SI unit for temperature, T is K.

The SI unit for the number of moles, n is mol.

$$\text{Therefore, the SI unit of quantity}\frac{pV^{2}T^{2}}{n}\text{is given by}\\=\frac{(\text{N}m^{\normalsize-2})(m^{3})^{2}(\text{K})^{2}}{(\text{mol})}\\=\frac{(\text{Nm}^{\normalsize-2})(m^{3})^{2}(\text{K})^{2}}{(\text{mol})}\\=\frac{\text{Nm}^{\normalsize-2}m^{6}\text{k}^{2}}{\text{mol}}$$

= Nm⁴ K² mol^–1

$$\text{Hence, the SI unit for quantity}\space\frac{\text{pV}^{2}T^{2}}{n}\space\text{is Nm}^4\text{K}^2 \text{mol}^{–1}.$$

5.21. In terms of Charles’ law explain why –273 °C is the lowest possible temperature.

Ans. According to the Charles law,

$$V_t=\bigg(V_0+\frac{V_0×t}{273}\bigg)=\text{V}_0\bigg(1+\frac{t}{273}\bigg)$$

where, V_t = volume of gas at temperature t°C

V₀ = volume of gas at 0°C

When t = – 273 °C

$$\text{V}_{-273\degree\text{C}}=\text{V}_0\bigg[1+\frac{(-273)}{(273)}\bigg]=\text{V}_0(1-1)=0.$$

Thus, according to Charles’ law, the volume of any gas would become zero at –273 °C. The lowest hypothetical temperature (–273°C) at which the volume of gas become zero is called absolute zero temperature. Any temperature below –273 °C would result in negative value which is physically impossible. Therefore, –273 °C is the lowest possible temperature.

5.22. Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C respectively. Which of these has stronger intermolecular forces and why?

Ans. Higher the critical temperature, more easily the gas can be liquefied. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, CO₂ has stronger intermolecular forces than CH₄.

5.23. Explain the physical significance of van der Waals parameters

Ans. The physical significance of Van der Waal’s parameter are as follows:

Van der Waal’s constant ‘a’ : The value of Van der waal’s constant ‘a’ is a measure of the magnitude of the intermolecular forces of attraction among the molecules of a gas. Greater the value of ‘a’, larger are the intermolecular forces of attraction.

Van der Waal’s constant ‘b’ : The value of Van der Waal’s constant ‘b’ is a measure of the effective size of the gas molecules. Its value is equal to four times the actual volume of the gas molecules. It is called as co-volume or excluded volume.