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

## NCERT Solutions for Class 11 Chemistry Chapter 5 Free PDF Download

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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 (V_{1}) = 500 dm^{3}

Initial pressure of a gas (P_{1}) = 1 bar at 30 °C.

Final volume of a gas (V_{2}) = 200 dm^{3}

Final pressure of a gas (P_{2}) = ? at 30 °C.

Since the temperature is kept constant, the final pressure (P_{2}) can be calculated using Boyle’s Law.

Thus,

According to Boyle’s law,

P_{1}V_{1} = P_{2}V_{2}

⇒ 1 bar × 500 dm^{3} = P_{2} × 200 dm^{3}

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

_{2}= 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 (V_{1}) = 120 mL

Initial pressure of a gas (P_{1}) = 1.2 bar at 35 °C.

Final volume of a gas (V_{2}) = 180 mL

Final pressure of a gas (P_{2}) = ? at 35 °C.

Since the temperature is kept constant, the final pressure (P_{2}) can be calculated using Boyle’s law.

Thus,

According to Boyle’s law,

P_{1}V_{1} = P_{2}V_{2}

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

P_{2} = 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, M_{1} and p_{1} are the mass and pressure of the oxide respectively.

Density of dinitrogen gas (d_{2}) is given by,

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

Where, M_{2} and p_{2} are the mass and pressure of the oxide respectively.

According to the question, it is given:

p_{1} = 2 bar

p_{2} = 5 bar

Molecular mass of nitrogen, M_{2} = 28 g/mol

Molecular mass of oxide, M_{1} = ?

Since,

d_{1} = d_{2}

∴ M_{1}p_{1} = M_{2}p_{2}

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

= 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_{A}V = n_{A}RT ...(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_{B}V = n_{B}RT ...(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 + 2H_{2}O → 2NaAlO_{2} + 3H_{2}

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 H_{2} = 186.7 mL

S.T.P. condition V_{1} = 186.7 mL, V_{2} = ?

p_{1} = 1.013 bar, p_{2} = 1 bar

T_{1} = 0 + 273 = 273 K, T_{2} = 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 dm^{3}, T = 27 + 273 = 300 K

R = 8.314 × 10^{–2} dm^{3} 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,

CH_{4} = 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,

CO_{2} = (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=PCH_{4}+PCO_{2}

= (0.5543 + 0.2771) bar = 0.8314 bar

= 0.8314 × 10^{5} Pa = 8.314 × 10^{4} Pa.

Hence, the pressure exerted will be 8.314 × 10^{4} 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 H_{2} and O_{2} be PH_{2} and PO_{2}, respectively.

The partial pressure of H_{2} in 1 L vessel,

Given P_{1} = 0.8 bar, V_{1} = 0.5 L, P_{2} = ? and V_{2 }= 1 L

According to Boyle’s law,

P_{1}V_{1} = P_{2}V_{2}

⇒ 0.8 bar × 0.5 L = P_{2} × 1 L

Hence, P_{2}= 0.40 bar, i.e., PH_{2} = 0.40 bar

The partial pressure of O_{2} in 1 L vessel,

Given P_{1}′ = 0.7 bar, V_{1}′ = 2 L

P_{2}′ = ? and V_{2}′ = 1 L

According to Boyle’s law,

P′_{1}V′_{1 }= P′_{2}V′_{2}

⇒ 0.7 bar × 2 L = P_{2} × 1 L

Hence, P′_{2} = 1.4 bar, i.e., P_{O2}

= 1.4 bar

Thus, total pressure will be,

P_{total} = PH_{2 }+ PO_{2}

= [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/dm ^{3 }at **

**27 °C at 2 bar pressure. What will be its density at STP?**

**Ans.** Given,

d_{1} = 5.46 g/dm^{3}

p_{1} = 2 bar

T_{1} = 27°C = (27 + 273) K = 300 K

p_{2} = 1 bar

T_{2} = 273 K

d_{2}= ?

The density (d_{2}) 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 cm^{3}

Then, the volume of air in the flask at 27°C =

V cm^{3}

Now,

V_{1} = V

V_{2} = ?

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,

T_{1} = 300 K and T_{2} = 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 dm ^{3} at 3.32 bar. (R = 0.083 bar dm^{3} K^{–1} mol^{–1}).**

**Ans.** Given,

n = 4 mol, V = 5 dm^{3}, P = 3.32 bar, R = 0.083 bar dm^{3} K^{–1} mol^{–1}

Using ideal gas equation,

PV = nRT

3.32 bar × 5 dm^{3} = 4 × 0.083 bar dm^{3} 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 (N_{2}) = 2 × 14 = 28 g

According to mole concept,

The number of molecules of N_{2} in 28 g

= 6.023 × 10^{23} molecules

The number of molecules of N2 in 1.4 g

$$=\frac{6.023×10^{23}×1.4}{28}$$

= 3.011 × 10^{22} molecules

Atomic number of nitrogen (N) = 7

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

Number of electrons in 1 molecule of N_{2}

= 7 × 2 = 14

Number of electrons in 3.01 × 10^{22} molecules of N_{2}

= (14 × 3.01 × 10^{22})

= 4.214 × 10^{23}

Hence, the total number of electrons present are 4.214 × 10^{23}.

**5.14. How much time would it take to distribute one Avogadro number of wheat grains, if 10 ^{10} grains are distributed each second ?**

**Ans.** One Avogadro’s number = 6.023 × 10^{23} particles

One Avogadro number of wheat grains

= 6.023 ×10^{23}grains

According to the question,

Time required to distribute 1010 grains = 1 s

Time required to distribute 6.02 × 10^{23} 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 × 10^{5} years or 1.909 × 10^{6} years

Hence, time taken would be 1.909 × 10^{6} 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 dm ^{3} K^{–1} mol^{–1}.**

**Ans.** Given,

V = 1 dm^{3}, T = 27 °C = 300 K, R = 0.083 bar dm^{3} K^{–1}mol^{–1}

Mass of dioxygen (O_{2}) = 8 g

Thus, Number of moles of O_{2} (nO_{2})

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

Thus, number of moles of H_{2} (n_{H2})

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

Mass of dihydrogen (H_{2}) = 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 m^{3}

The mass of displaced air

= 4190.5 m^{3} × 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^{–3}kg mol^{–1}

p = 1.66 bar

V = Volume of the balloon

= 4190.5 m^{3}

R = 0.083 bar dm^{3}K^{–1}mol^{–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 CO _{2} 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^{–1}mol^{–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 CO_{2} 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 H_{2} gas, |
For unknown gas, |

p_{1}= p |
p_{2} = p |

V_{1} = V |
V_{2} = V |

T_{1} = 17+273 = 290 K |
T_{2} = 95 + 273 = 368 K |

m_{1}=0.184 g |
m_{2} = 2.9 |

M_{1} = 2 g mol^{–1} |
M_{2} = ? |

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 **_{p}V^{2}T^{2}/n ?

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

The SI unit for volume, V is m^{3}.

The SI unit for temperature, T is K.

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

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

= Nm^{4} K^{2} 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

V_{0} = 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, CO_{2} has stronger intermolecular forces than CH_{4}.

**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**.