NCERT Solutions for Class 11 Chemistry Chapter 5 - States Of Matter

NCERT Exercise

1. What will be the minimum pressure required to compress 500 $\text{d}{{\text{m}}^{\text{3}}}$of air at 1 bar to 200 $\text{d}{{\text{m}}^{\text{3}}}$ at $\text{3}{{\text{0}}^{\text{o}}}\text{C}$?  

Ans: Given, 

Initial pressure,${{p}_{1}}$ = 1 bar 

Initial volume, ${{V}_{1}}$ = 500 $d{{m}^{3}}$  

Final volume,${{V}_{2}}$ = 200 $d{{m}^{3}}$ 

Since the temperature remains constant, the final pressure (${{p}_{2}}$) can be calculated using Boyle’s law. 

According to Boyle’s law, 

\[{{p}_{1}}{{V}_{1}}={{p}_{2}}{{V}_{2}}\] 

\[\Rightarrow {{p}_{2}}=\frac{{{p}_{1}}{{V}_{1}}}{{{V}_{2}}}\] 

\[=\frac{1\times 500}{200}bar\] 

\[=\text{ }2.5\text{ }bar\] 

Therefore, the minimum pressure required is 2.5 bar.

2. A vessel of 120 mL capacity contains a certain amount of gas at $\text{3}{{\text{5}}^{\text{o}}}\text{C}$ and 1.2 bar pressure. The gas is transferred to an answer vessel of volume 180 mL at $\text{3}{{\text{5}}^{\text{o}}}\text{C}$. What would be its pressure? 

Ans: Given, 

Initial pressure,${{p}_{1}}$  = 1.2 bar 

Initial volume,${{V}_{1}}$ = 120 mL 

Final volume,${{V}_{2}}$  = 180 mL 

Since the temperature remains constant, the final pressure (${{p}_{2}}$) can be calculated using Boyle’s law. 

According to Boyle’s law,

\[{{p}_{1}}{{V}_{1}}={{p}_{2}}{{V}_{2}}\] 

\[\Rightarrow {{p}_{2}}=\frac{{{p}_{1}}{{V}_{1}}}{{{V}_{2}}}\] 

\[=\frac{1.2\times 120}{180}bar\] 

\[=\text{ }0.8\text{ }bar\] 

Therefore, the pressure would be 0.8 bar.

3. Using the equation of state pV = nRT shows that at a given temperature density of a gas is proportional to gas pressure ep. 

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, 

\[\frac{n}{V}=\frac{p}{RT}\] 

Replacing n with$\frac{m}{M}$  , we have 

\[\frac{m}{MV}=\frac{p}{RT}......(ii)\] 

Where, 

m = Mass of gas 

M = Molar mass of gas 

But, $\frac{m}{V}=d$ (d = density of gas) 

Thus, from equation (ii), we have 

\[\frac{d}{M}=\frac{p}{RT}\] 

\[\Rightarrow d=\left( \frac{M}{RT} \right)p\] 

Molar mass (M) of gas is always constant and therefore, at constant temperature (T),$\frac{M}{RT}$ = constant, 

d = (constant) p 

$\Rightarrow d\propto p$ 

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

4. At ${{\text{0}}^{\text{o}}}\text{C}$ , the density of certain oxide of a gas at 2 bar is the same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?  

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

\[d=\frac{Mp}{RT}\] 

Now, density of oxide (${{d}_{1}}$) is given by, 

\[{{d}_{1}}=\frac{{{M}_{1}}{{p}_{1}}}{RT}\] 

Where, ${{M}_{1}}$  and ${{p}_{1}}$ are the mass and pressure of the oxide respectively. 

Density of dinitrogen gas (${{d}_{2}}$) is given by, 

\[{{d}_{2}}=\frac{{{M}_{2}}{{p}_{2}}}{RT}\] 

Where, ${{M}_{2}}$  and ${{p}_{2}}$ are the mass and pressure of the oxide respectively. 

According to the given question, 

\[{{d}_{1}}={{d}_{2}}\] 

Therefore,

\[{{M}_{1}}{{p}_{1}}={{M}_{2}}{{p}_{2}}\] 

Given, 

${{p}_{1}}$  = 2 bar 

${{p}_{2}}$  = 5 bar 

Molecular mass of nitrogen, ${{M}_{2}}$  = 28 g/mol 

Now, ${{M}_{1}}=\frac{{{M}_{2}}{{p}_{2}}}{{{p}_{1}}}$ 

\[=\frac{28\times 5}{2}\] 

\[=70g/mol\] 

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

5. Pressure of 1 g of an ideal gas A $\text{2}{{\text{7}}^{\text{o}}}\text{C}$  is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses. 

Ans: For ideal gas A, the ideal gas equation is given by, 

\[{{p}_{A}}{{V}_{A}}=nRT....(i)\] 

Where, ${{p}_{A}}$ and ${{V}_{A}}$ represent the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is given by, 

\[{{p}_{B}}{{V}_{B}}=nRT....(ii)\] 

Where, ${{p}_{B}}$ and${{V}_{B}}$ 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 

\[{{p}_{A}}V=\frac{{{m}_{A}}}{{{M}_{A}}}RT\Rightarrow \frac{{{p}_{A}}{{M}_{A}}}{{{m}_{A}}}=\frac{RT}{V}......(iii)\] 

From equation (ii), we have 

\[{{p}_{B}}V=\frac{{{m}_{B}}}{{{M}_{B}}}RT\Rightarrow \frac{{{p}_{B}}{{M}_{B}}}{{{m}_{B}}}=\frac{RT}{V}......(iv)\] 

Where, ${{M}_{A}}$  and ${{M}_{B}}$  are the molecular masses of gases A and B respectively. 

Now, from equations (iii) and (iv), we have 

\[\frac{{{p}_{A}}{{M}_{A}}}{{{m}_{A}}}=\frac{{{p}_{B}}{{M}_{B}}}{{{m}_{B}}}.......(v)\] 

Given, 

${{m}_{A}}$ =1g

${{p}_{A}}$ =2bar

${{m}_{B}}$ = 2g

${{p}_{B}}$ = (3-2) = 1bar

(Since total pressure is 3 bar) 

Substituting these values in equation (v), we have 

\[\frac{2\times {{M}_{A}}}{1}=\frac{1\times {{M}_{B}}}{2}\] 

\[\Rightarrow 4{{M}_{A}}={{M}_{B}}\] 

Thus, a relationship between the molecular masses of A and B is given by $4{{M}_{A}}={{M}_{B}}$.

5. The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at $\text{2}{{\text{0}}^{\text{o}}}\text{C}$ and one bar will be released when 0.15g of aluminum reacts? 

Ans: The reaction of aluminum with caustic soda can be represented as: 

\[2Al+2NaOH+2{{H}_{2}}O\to 2NaAl{{O}_{2}}+3{{H}_{2}}\] 

At STP (273.15 K and 1 atm), 54 g (2 $\times $ 27 g) of Al gives 3$\times $ 22400 mL of ${{H}_{2}}$. 

Therefore, 0.15 g Al gives$\frac{3\times 22400\times 0.15}{54}mL$ of${{H}_{2}}$  i.e., 186.67 mL of ${{H}_{2}}$. 

At STP,

${{p}_{1}}$ = 1atm

${{V}_{1}}$ = 186.67 mL

${{T}_{1}}$ = 273.15 K

Let the volume of dihydrogen be ${{V}_{2}}$  at ${{p}_{2}}$  = 0.987 atm (since 1 bar = 0.987 atm) and ${{T}_{2}}$  = ${{20}^{o}}C$ = (273.15 + 20) K = 293.15 K. 

\[\frac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{p}_{2}}{{V}_{2}}}{{{T}_{2}}}\] 

\[{{V}_{2}}=\frac{{{p}_{1}}{{V}_{1}}{{T}_{2}}}{{{p}_{2}}{{T}_{1}}}\] 

\[=\frac{1\times 186.67\times 293.15}{0.987\times 273.15}\] 

\[=202.98mL\] 

$=203mL$ 

Therefore, 203 mL of dihydrogen will be released.

6. 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 $\text{d}{{\text{m}}^{\text{3}}}$  flask at $\text{2}{{\text{7}}^{\text{o}}}\text{C}$ ?

Ans: It is known that, 

$p=\frac{m}{M} \frac{RT}{V}$ 

For methane ($C{{H}_{4}}$),

\[{{p}_{C{{H}_{4}}}}=\frac{3.2}{16}\times \frac{8.314\times 300}{9\times {{10}^{-3}}} \]

 since \[9d{{m}^{3}}= 9 \times {{10}^{-3}}{{m}^{3}} \]

\[  {{27}^{o}}C=300K \] 

\[=5.543\times {{10}^{4}}Pa\]

For carbon dioxide ($C{{O}_{2}}$),

\[{{p}_{C{{O}_{2}}}}=\frac{4.4}{44}\times \frac{8.314\times 300}{9\times {{10}^{-3}}}\] 

\[=2.771\times {{10}^{4}}Pa\] 

The pressure exerted by the mixture can be obtained as: 

\[p={{p}_{C{{H}_{4}}}}+{{p}_{C{{O}_{2}}}}\] 

\[=(5.543\times {{10}^{4}}+2.771\times {{10}^{4}})Pa\] 

\[=8.314\times {{10}^{4}}Pa\] 

Hence, the total pressure exerted by the mixture is $=8.314\times {{10}^{4}}Pa$.

7. What will be the pressure of the gaseous mixture when 0.5 L of ${{\text{H}}_{\text{2}}}$  at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at $\text{2}{{\text{7}}^{\text{o}}}\text{C}$ ? 

Ans: Let the partial pressure of ${{H}_{2}}$in the vessel be ${{p}_{{{H}_{2}}}}$. 

Now, 

${{p}_{1}}$ = 0.8bar 

${{V}_{1}}$ = 0.5L

${{V}_{2}}$ = 1L

${{p}_{2}}={{p}_{{{H}_{2}}}}$= ?

It is known that,

\[{{p}_{1}}{{V}_{1}}={{p}_{2}}{{V}_{2}}\] 

\[\Rightarrow {{p}_{2}}=\frac{{{p}_{1}}{{V}_{1}}}{{{V}_{2}}}\] 

\[\Rightarrow {{p}_{{{H}_{2}}}}=\frac{0.8\times 0.5}{1}\] 

\[=0.4bar\]  

Now, let the partial pressure of ${{O}_{2}}$in the vessel be${{p}_{{{O}_{2}}}}$. 

${{p}_{1}}$ = 0.7bar

${{V}_{1}}$ = 2.0L

${{V}_{2}}$ = 1L

${{p}_{2}}={{p}_{{{O}_{2}}}}$ = ?

Since,

\[{{p}_{1}}{{V}_{1}}={{p}_{2}}{{V}_{2}}\] 

\[\Rightarrow {{p}_{2}}=\frac{{{p}_{1}}{{V}_{1}}}{{{V}_{2}}}\] 

\[\Rightarrow {{p}_{{{O}_{2}}}}=\frac{0.7\times 20}{1}\] 

\[=1.4bar\] 

Total pressure of the gas mixture in the vessel can be obtained as: 

\[{{p}_{total}}={{p}_{{{H}_{2}}}}+{{p}_{{{O}_{2}}}}\]

\[=0.4+1.4\] 

\[=1.8bar\]  

Hence, the total pressure of the gaseous mixture in the vessel is 1.8 bar. 

8. Density of a gas is found to be 5.46 $\text{g/d}{{\text{m}}^{\text{3}}}$ at $\text{2}{{\text{7}}^{\text{o}}}\text{C}$ at 2 bar pressure. What will be its density at STP?

Ans: Given, 

${{d}_{1}}$ = 5.46$g/d{{m}^{3}}$ 

${{p}_{1}}$ =2 bar

${{T}_{1}}={{27}^{o}}C$ = (27+273)K=300K

${{p}_{2}}=$ 1bar

${{T}_{2}}$ =273K

${{d}_{2}}$ =?

The density (${{d}_{2}}$) of the gas at STP can be calculated using the equation, 

\[d=\frac{Mp}{RT}\] 

\[\therefore \frac{{{d}_{1}}}{{{d}_{2}}}\frac{\frac{M{{p}_{1}}}{R{{T}_{1}}}}{\frac{M{{p}_{2}}}{R{{T}_{2}}}}\] 

\[\Rightarrow \frac{{{d}_{1}}}{{{d}_{2}}}=\frac{{{p}_{1}}{{T}_{2}}}{{{p}_{2}}{{T}_{1}}}\] 

\[\Rightarrow {{d}_{2}}=\frac{{{p}_{2}}{{T}_{1}}{{d}_{1}}}{{{p}_{1}}{{T}_{2}}}\] 

\[=\frac{1\times 300\times 5.46}{2\times 273}\] 

\[=3gd{{m}^{-3}}\] 

Hence, the density of the gas at STP will be 3 g$d{{m}^{-3}}$.

9. 34.05 mL of phosphorus vapour weighs 0.0625 g at $\text{54}{{\text{6}}^{\text{o}}}\text{C}$ and 0.1 bar pressure. What is the molar mass of phosphorus? 

Ans: Given, 

p = 0.1 bar 

V = 34.05 mL = $34.05\times {{10}^{-3}}L=34.05\times {{10}^{-3}}d{{m}^{-3}}$  

R =0.083 bar $d{{m}^{3}}{{K}^{-1}}mo{{l}^{-1}}$ 

T = ${{546}^{o}}C$ = (546 + 273) K = 819 K 

The number of moles (n) can be calculated using the ideal gas equation as:

\[pV=nRT\] 

\[\Rightarrow n=\frac{pV}{RT}\] 

\[=\frac{0.1\times 34.05\times {{10}^{-3}}}{0.083\times 819}\] 

\[=5.01\times {{10}^{-5}}mol\] 

Therefore, molar mass of phosphorus =$\frac{0.0625}{5.01\times {{10}^{-5}}}=1247.5gmo{{l}^{-1}}$ 

Hence, the molar mass of phosphorus is $1247.5\,gmo{{l}^{-1}}$.

10. A student forgot to add the reaction mixture to the round bottomed flask at $\text{2}{{\text{7}}^{\text{o}}}\text{C}$  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 $\text{47}{{\text{7}}^{\text{o}}}\text{C}$ . What fraction of air would have been expelled out? 

Ans: Let the volume of the round bottomed flask be V. 

Then, the volume of air inside the flask at ${{27}^{o}}C$is V. 

Now, 

${{V}_{1}}$  = V 

${{T}_{1}}$  = ${{27}^{o}}C$ = 300 K 

${{V}_{2}}$ = ? 

${{T}_{2}}$ = ${{477}^{o}}C$  = 750 K 

According to Charles’s law, 

\[\frac{{{V}_{1}}}{{{T}_{1}}}=\frac{{{V}_{2}}}{{{T}_{2}}}\] 

\[\Rightarrow {{V}_{2}}=\frac{{{V}_{1}}{{T}_{2}}}{{{T}_{1}}}\] 

\[=\frac{750V}{300}\] 

\[=2.5V\] 

Therefore, volume of air expelled out = 2.5 V – V = 1.5 V 

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

11. Calculate the temperature of 4.0 mol of gas occupying 5 $\text{d}{{\text{m}}^{\text{3}}}$  at 3.32 bar. (R = 0.083 bar$\text{d}{{\text{m}}^{\text{3}}}{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}$). 

Ans: Given, 

n = 4.0 mol 

V = 5 $d{{m}^{3}}$ 

p = 3.32 bar 

R = 0.083 bar $d{{m}^{3}}{{K}^{-1}}mo{{l}^{-1}}$ 

The temperature (T) can be calculated using the ideal gas equation as: 

\[pV=nRT\] 

\[\Rightarrow T=\frac{pV}{nR}\] 

\[=\frac{3.32\times 5}{4\times 0.083}\] 

\[=50K\] 

Hence, the required temperature is 50 K. 

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

Ans: Molar mass of dinitrogen (${{N}_{2}}$) = 28 $g\,mo{{l}^{-1}}$ 

Thus, 1.4 g of ${{N}_{2}}$ $\frac{1.4}{28}$ = 0.05 mol

= 0.05$\times 6.02\times {{10}^{23}}$ number of molecules 

= 3.01$\times {{10}^{23}}$  number of molecules 

Now, 1 molecule of ${{N}_{2}}$contains 14 electrons. 

Therefore, 3.01$\times {{10}^{23}}$  molecules of ${{N}_{2}}$contains = $14\times 3.01\times {{10}^{23}}$ = $4.214\times {{10}^{23}}$ electrons.

13. How much time would it take to distribute one Avogadro number of wheat grains, if $\text{1}{{\text{0}}^{\text{10}}}$ grains are distributed each second? 

Ans: Avogadro number = $6.02\times {{10}^{23}}$ 

Thus, time required

\[=\frac{6.02\times {{10}^{23}}}{{{10}^{10}}}s\] 

\[=6.02\times {{10}^{23}}s\] 

\[=\frac{6.02\times {{10}^{23}}}{60\times 60\times 24\times 365}years\] 

\[=1.909\times {{10}^{6}}years\] 

Hence, the time taken would be$=1.909\times {{10}^{6}}years$.

14. Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 $\text{d}{{\text{m}}^{\text{3}}}$ at $\text{2}{{\text{7}}^{\text{o}}}\text{C}$ . (R = 0.083 bar$\text{d}{{\text{m}}^{\text{3}}}{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}$ ). 

Ans: Given, 

Mass of dioxygen (${{O}_{2}}$) = 8 g 

Thus, number of moles of ${{O}_{2}}$ =$\frac{8}{32}$ = 0.25 mole

Mass of dihydrogen (${{H}_{2}}$) = 4 g

${{H}_{2}}=\frac{4}{2}$ = 2mole 

Therefore, total number of moles in the mixture = 0.25 + 2 2.25 mole 

Given, 

V = 1 $d{{m}^{3}}$ 

n = 2.25 mol 

R = 0.083 bar $d{{m}^{3}}{{K}^{-1}}mo{{l}^{-1}}$ 

T = ${{27}^{o}}C$  = 300 K 

Total pressure (p) can be calculated as: 

\[pV=nRT\]

\[\Rightarrow p=\frac{nRT}{V}\] 

\[=\frac{225\times 0.083\times 300}{1}\] 

\[=56.025bar\]  

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

15. Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar $\text{2}{{\text{7}}^{\text{o}}}\text{C}$ . (Density of air = 1.2$\text{kg}\,{{\text{m}}^{\text{-3}}}$ . And R = 0.083 bar$\text{d}{{\text{m}}^{\text{3}}}{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}$ ). 

Ans: Given, 

Radius of the balloon, r = 10 m 

Therefore, Volume of the balloon =$\frac{4}{3}\pi {{r}^{3}}$ 

\[=\frac{4}{3}\times \frac{22}{7}\times {{10}^{23}}\] 

\[=4190.5{{m}^{3}}(approx)\] 

Thus, the volume of the displaced air is 4190.5 ${{m}^{3}}$. 

Given, 

Density of air = 1.2 $kg\,{{m}^{-3}}$  

Then, mass of displaced air = 4190.5${{m}^{3}}$  $\times $ 1.2$kg\,{{m}^{-3}}$  

= 5028.6 kg 

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

\[m=\frac{MpV}{RT}\] 

Here, 

M = 4$\times {{10}^{-3}}kgmo{{l}^{-1}}$ 

p = 1.66 bar 

V = Volume of the balloon 

= 4190.5 ${{m}^{3}}$ 

R = 0.083 bar $d{{m}^{3}}{{K}^{-1}}mo{{l}^{-1}}$ 

T = ${{27}^{o}}C$  = 300 K 

Then, 

\[m=\frac{4\times {{10}^{-3}}\times 1.66\times 4190.5\times {{10}^{3}}}{0.083\times 300}\] 

\[=1117.5kg(approx)\] 

Now, total mass of the balloon filled with helium = (100 + 1117.5) kg 

= 1217.5 kg 

Hence, payload = (5028.6 – 1217.5) kg 

=3811.1 kg 

Hence, the payload of the balloon is 3811.1 kg.

16. Calculate the volume occupied by 8.8 g of $\text{C}{{\text{O}}_{\text{2}}}$ at $\text{31}\text{.}{{\text{1}}^{\text{o}}}\text{C}$ and 1 bar pressure. R = 0.083 bar L${{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}$. 

Ans: It is known that,

\[pV=\frac{m}{N}RT\] 

\[\Rightarrow V=\frac{mRT}{Mp}\] 

Here, 

m = 8.8 g 

R = 0.083 bar L${{K}^{-1}}mo{{l}^{-1}}$  

T = ${{31.1}^{o}}C$ = 304.1 K 

M = 44 g 

p = 1 bar 

Thus, Volume (V) $=\frac{8.8\times 0.083\times 304.1}{44\times 1}$ 

\[=\text{ }5.04806\text{ }L\] 

\[=\text{ }5.05\text{ }L\] 

Hence, the volume occupied is 5.05 L. 

17. 2.9 g of gas at $\text{9}{{\text{5}}^{\text{o}}}\text{C}$  occupied the same volume as 0.184 g of dihydrogen at $\text{1}{{\text{7}}^{\text{o}}}\text{C}$ , at the same pressure. What is the molar mass of the gas? 

Ans: Volume (V) occupied by dihydrogen is given by, 

\[V=\frac{m}{M}\frac{RT}{p}\] 

\[=\frac{0.184}{2}\times \frac{R\times 290}{p}\] 

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

\[V=\frac{m}{M}\frac{RT}{p}\] 

\[=\frac{2.9}{M}\times \frac{R\times 368}{p}\] 

According to the equation,

\[\frac{0.184}{2}\times \frac{R\times 290}{p}=\frac{2.9}{M}\times \frac{R\times 368}{p}\] 

\[\Rightarrow \frac{0.184\times 290}{2}=\frac{2.9\times 368}{M}\] 

\[M=\frac{2.9\times 368\times 2}{0.184\times 290}\] 

\[=40gmo{{l}^{-1}}\] 

Hence, the molar mass of the gas is$40\,g\,mo{{l}^{-1}}$.

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

Ans: 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 moles and the number of moles of dioxygen,${{n}_{{{O}_{2}}}}=\frac{80}{32}$ =2.5 moles. 

Given, 

Total pressure of the mixture, ${{P}_{total}}$ = 1bar 

Then, partial pressure of dihydrogen, \[{{p}_{{{H}_{2}}}}=\frac{{{n}_{{{H}_{2}}}}}{{{n}_{{{H}_{2}}}}+{{n}_{{{O}_{2}}}}}\times {{P}_{total}}\] 

\[=\frac{10}{10+2.5}\times 1\] 

\[=0.8bar\] 

Hence, the partial pressure of dihydrogen is 0.8 bar. 

19. What would be the SI units for the quantity $\text{p}{{\text{V}}^{\text{2}}}{{\text{T}}^{\text{2}}}\text{/n}$? 

Ans: The SI units for pressure, p is$N{{m}^{-2}}$ . 

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.

Therefore, the SI unit for quantity $\frac{p{{V}^{2}}{{T}^{2}}}{n}$ is given by, 

\[=\frac{\left( N{{m}^{-2}} \right){{\left( {{m}^{3}} \right)}^{2}}{{\left( K \right)}^{2}}}{mol}\] 

\[=N{{m}^{4}}{{K}^{2}}mo{{l}^{-1}}\] 

20. In terms of Charles’ law explain why $\text{-27}{{\text{3}}^{\text{o}}}\text{C}$ is the lowest possible temperature. 

Ans: Charles’s law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature. 

(Image Will Be Updated Soon)

It was found that for all gases (at any given pressure), the plots of volume vs. temperature (in$^{o}C$ ) is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis at$-{{273}^{o}}C$ . In other words, the volume of any gas at ${{273}^{o}}C$ is zero. This is because all gases get liquefied before reaching a temperature of${{273}^{o}}C$ . Hence, it can be concluded that $-{{273}^{o}}C$is the lowest possible temperature.

21. Critical temperature for carbon dioxide and methane are $\text{31}\text{.}{{\text{1}}^{\text{o}}}\text{C}$ and –$\text{81}\text{.}{{\text{9}}^{\text{o}}}\text{C}$ respectively. Which of these has stronger intermolecular forces and why? 

Ans: Higher is the critical temperature of a gas, easier is its liquefaction. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case of$C{{O}_{2}}$ .

22. Explain the physical significance of Van der Waals parameters. 

Ans: The Van der waals equation is an equation of state for a fluid composed of particles that have a non-zero volume and a pair wise attractive inter-particle force ( Van der waals force) The equation is :

\[\left( p+\frac{{{n}^{2}}a}{{{V}^{2}}} \right)\left( V-nb \right)=nRT\] 

Physical significance of ‘a’: 

‘a’ is a measure of the magnitude of intermolecular attractive forces within a gas. 

Physical significance of ‘b’: 

‘b’ is a measure of the volume of a gas molecule. 

V is the total volume of the container containing the fluid.

Let us Quickly Give a Glance at the Properties of the Four States of Matter

Solid: In solids, the atoms, molecules, and ions are closely packed together and so the particles can vibrate at their own position but are not free to move. The volume and the shape of a solid can change when an external force is applied to them.

Liquid: The atoms, ions, and molecules are incompressible in liquid. They do not depend on pressure. They have a fixed volume when the pressure and temperature are constant. When solid is exposed to very high temperatures they transform into the liquid state, subject to the properties of pressure.

Gas: The intermolecular force of attraction between the interacting particles has enough kinetic energy that changes the molecular forces to zero. The intermolecular space between the molecules is very large and so a liquid can be transformed into a gaseous state.

Plasma: Plasma is a state of matter in which the ions are highly electrically conducive. The magnetic fields and currents produced by them dominate the behaviour of the matter.

So we see that the behaviour of matter in different states of matter is governed by factors like temperature, pressure, mass, and volume. The chemical properties of a substance do not change with the change of state but chemical reactions depend on the physical state of the matter.

Chemistry NCERT Solutions for Class 11 - Chapter wise PDFs

• Chapter 1 - Some Basic Concepts of Chemistry

• Chapter 2 - Structure of Atom

• Chapter 3 - Classification of Elements and Periodicity in Properties

• Chapter 4 - Chemical Bonding and Molecular Structure

• Chapter 5 - States of Matter

• Chapter 6 - Thermodynamics

• Chapter 7 - Equilibrium

• Chapter 8 - Redox Reactions

• Chapter 9 - Hydrogen

• Chapter 10 - The s-Block Elements

• Chapter 11 - The p-Block Elements

• Chapter 12 - Organic Chemistry - Some Basic Principles and Techniques

• Chapter 13 - Hydrocarbons

• Chapter 14 - Environmental Chemistry

Intermolecular Forces

The intermolecular forces interact between the particles of matter. These forces are different from electrostatic forces that exist between two oppositely charged ions.

Different types of Intermolecular forces are:

1. Dispersion Forces or London Forces

2. Dipole - Dipole Forces

3. Dipole - Induced Dipole Forces

4. Hydrogen Bond

Thermal Energy

When the motion of atoms or molecules in a body causes an energy to arise is known as Thermal Energy. It is a quantity of the average kinetic energy of the particles. It increases with an increase in temperature.

Intermolecular Forces Vs Thermal Interactions

As we have just learned that intermolecular forces in a substance tend to keep the molecules together but the thermal energy of the particles tends to keep them apart. Therefore, the three states of matter are the resultant of the balance between the intermolecular forces and thermal energy of the molecules.

Gaseous State

In the Gaseous state of matter the physical properties of the molecules are:

• The molecules in the gases are highly compressible.

• Gases exert pressure in all directions.

• The density in the gases is much lower than the solids and liquids.

• Gases do not have definite volume and shape. They take the shape of a container.

• Gases mix completely in all proportions.

The forces of interaction between gas molecules are negligible. Some laws that were discovered as a result during some experimental studies govern the behaviour of the gaseous molecules.

Boyle’s Law

This law states that under isothermal conditions, the pressure of a fixed amount of a gas is inversely proportional to its volume.

P1V1 = P2V2 (at constant T)

where P = Pressure, V= Volume and T = Temperature.

If a fixed amount of gas at a constant temperature T occupying volume V1 at pressure P1 , expands such that volume becomes V2 and pressure becomes P2.

Charles's Law

Charle’s law is a direct relationship between volume and absolute temperature under isobaric condition. It states that for a fixed amount of gas at a constant temperature, the volume of a gas increases with the increase in temperature and decreases on cooling. The volume of the gas is directly proportional to its absolute temperature.

V1  =  V2

T1  = V2     (at constant p)

Where V = Volume, T = Temperature, P = Pressure.

Gay Lussac’s Law

This law states that the pressure of a fixed amount of a gas is directly proportional to its absolute temperature at constant volume.

P ∝ T   (at constant v)

where P = Pressure, T = Temperature, V = Volume.

Avogadro Law

Avogadro Law states that equal volumes of all gases under the same conditions of temperature and pressure contain equal numbers of molecules.

V ∝ n (at constant T and p)

where V = Volume, n = number of molecules, T = Temperature, P = Pressure.

Ideal Gas Equation

The three laws – Boyle’s law, Gay Lussac’s Law, and Avogadro Law are combined together to form a single equation, which is known as the ideal gas equation.

pV = nRT

where p = Pressure, V = Volume, n = number of molecules, R = gas constant, T = Temperature.

Dalton’s Law of Partial Pressure

This law states that total pressure exerted by a mixture of non-reacting gases is equal to the sum of partial pressures exerted by them.

Note: In a mixture of gases, the pressure exerted by the individual gas is called partial pressure.

Ptotal = p1 + p2 + p3 + ……..  where, p1, p2, ……. are partial pressures.

Kinetic Theory of Gases

• Gases contain a large number of identical molecules.

• The molecules in gases are far apart from each other and so the actual volume occupied by the molecules is negligible. This is why gases can be compressed.

• There is no force of attraction between the molecules of a gas at ordinary temperature and pressure. This is why gases can expand and occupy space.

• The gas particles move in all directions in a straight line. They collide with each other during their random motion.

• The collision of the gas particles is elastic in nature and so there is no loss of kinetic energy.

• The speed of every molecule in the gases is different and their speed keeps changing. So their energy also keeps changing.