NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter

Ans : Given, 
Initial pressure, \(P_{1}\) = 1 bar 
Initial volume, \(V_{1}\) = 500  \(\mathrm{dm}^{3}\)
Final volume, \(V_{2}\) = 200 \(\mathrm{dm}^{3}\) 
Since the temperature remains constant, the final pressure \((P_{2})\) can be calculated using Boyle's law. 
According to Boyle's law, 
\(\begin{aligned} p_{1} V_{1} &=p_{2} V_{2} \\ \Rightarrow p_{2} &=\frac{p_{1} V_{1}}{V_{2}} \\ &=\frac{1 \times 500}{200} \mathrm{bar} \\ &=2.5 \mathrm{bar} \end{aligned}\)
Therefore, the minimum pressure required is 2.5 bar. 

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,
\(\begin{aligned} p_{1} V_{1} &=p_{2} V_{2} \\ p_{2} &=\frac{p_{1} V_{1}}{V_{2}} \\ &=\frac{1.2 \times 120}{180} \text { bar } \\ &=0.80 \\ \text { Therefore, the pressure would be } 0.8 \text { bar. } \end{aligned}\) 

Ans : The equation of state is given by, 
\(\rho V=n R T \ldots \ldots \ldots\)
Where, 
P \(\rightarrow\) Pressure of gas  
V \(\rightarrow\) Volume of gas 
n \(\rightarrow\) Number of moles of gas 
R \(\rightarrow\) Gas constant 
T \(\rightarrow\) Temperature of gas 
From equation (i) we have, 
\(\begin{array}{l}{\frac{n}{V}=\frac{p}{\mathrm{R} T}} \\ {\text { Replacing } n \text { with } \frac{m}{M}, \text { we have }} \\ {\frac{m}{M V}=\frac{p}{\mathrm{R} T} \ldots \ldots \ldots(\text { ii })} \\ {\text { Where, }} \\ {m \rightarrow \text { Mass of gas }} \\ {M \rightarrow \text { Molar mass of gas }}\end{array}\)
\(\begin{array}{l}{\frac{m}{V}=d} \\ {\text { Thus, from equation (ii), we have }} \\ {\frac{d}{M}=\frac{p}{\mathrm{R} T}} \\ {\Rightarrow d=\left(\frac{M}{\mathrm{R} T}\right) P}\end{array}\)
\(\begin{array}{l}{\text { Molar mass }(M) \text { of a gas is always constant and therefore, at constant temperature }} \\ {(T), \frac{M}{R T}=\text { constant. }} \\ {d=(\text { constant }) p} \\ {\Rightarrow d \propto p} \\ {\text { Hence, at a given temperature, the density (d) of gas is proportional to its pressure (p) }}\end{array}\) 

Ans : \(\begin{array}{l}{\text { Density (d) of the substance at temperature (T) can be given by the expression, }} \\ {\quad d=\frac{M p}{R T}} \\ {\text { Now, density of oxide }\left(d_{1}\right) \text { is given by, }}\end{array}\)
\(\begin{array}{l}{d_{1}=\frac{M_{1} p_{1}}{R T}} \\ {\text { Where, } M_{2} \text { and } p_{1} \text { are the mass and pressure of the oxide respectively. }} \\ {\text { Density of dinitrogen gas }\left(d_{2}\right) \text { ls given by, }} \\ {d_{2}=\frac{M_{2} p_{2}}{R T}}\end{array}\)
\(\begin{array}{l}{\text { Where, } M_{2} \text { and } p_{2} \text { are the mass and pressure of the oxide respectively. }} \\ {\text { According to the given question, }} \\ {d_{1}=d_{2}} \\ {\therefore M_{1} p_{1}=M_{2} p_{2}} \\ {\text { Given, }} \\ {p_{1}=2 b a r} \\ {p_{2}=5 \text { bar }}\end{array}\)
\(\begin{aligned} \text { Molecular } & \text { mass of nitrogen, } M_{2}=28 \mathrm{g} / \mathrm{mol} \\ \mathrm{Now}, M_{1} &=\frac{M_{2} p_{2}}{p_{1}} \\ &=\frac{28 \times 5}{2} \\ &=70 \mathrm{g} / \mathrm{mol} \\ \text { Hence, the molecular mass of the oxlde is } 70 \mathrm{g} / \mathrm{mol} \end{aligned}\) 

Ans : \(\begin{array}{l}{\text { For ideal gas } A, \text { the ideal gas equation is given by, }} \\ {p_{\mathrm{A}} V=n_{\mathrm{A}} \mathrm{RT} \ldots \ldots \text { (i) }}\end{array}\)
Where, \(p_{\mathrm{A}} \text { and } \mathrm{n}_{\mathrm{B}}\) represent the pressure and number of moles of gas A. 
For ideal gas B, the ideal gas equation is given by,
\(p_{B} V=n_{B} \mathrm{RT} \ldots \ldots .\)
Where, p and n, 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
\(p_{A} V=\frac{m_{A}}{\mathbf{M}_{A}} \mathrm{R} T \Rightarrow \frac{p_{A} \mathrm{M}_{A}}{m_{A}}=\frac{\mathrm{RT}}{V} \ldots \ldots \ldots\)
From equation (ii), we have 
\(p_{\mathrm{B}} V=\frac{m_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}} \mathrm{R} T \Rightarrow \frac{p_{\mathrm{B}} \mathrm{M}_{\mathrm{B}}}{m_{\mathrm{B}}}=\frac{\mathrm{R} T}{V} \ldots \ldots \ldots\)
Where, \(\mathrm{M}_{\mathrm{A}} \text { and } \mathrm{M}_{\mathrm{B}}\) are the molecular masses of gases A and B respectively. 
Now, from equations (iii) and (iv) we have 
\(\frac{p_{A} \mathbf{M}_{A}}{m_{A}}=\frac{p_{g} \mathbf{M}_{B}}{m_{B}} \ldots \ldots \ldots(v)\)
Given, 
\(\begin{array}{l}{m_{A}=1 \mathrm{g}} \\ {p_{A}=2 \text { bar }} \\ {m_{B}=2 \mathrm{g}} \\ {p_{B}=(3-2)=1 \mathrm{bar}}\end{array}\)
\(\begin{array}{l}{\text { (Since total pressure is } 3 \text { bar) }} \\ {\text { Substituting these values in equation (v), we have }} \\ {\frac{2 \times \mathrm{M}_{A}}{1}=\frac{1 \times \mathrm{M}_{B}}{2}} \\ {\Rightarrow4\mathrm{M}_{A}=\mathrm{M}_{B}}\end{array}\)
\(\begin{array}{l}{\text { Thus, a relationship between the molecular masses of } A \text { and } B \text { is given by }} \\ {4 M_{A}=M_{B}}\end{array}\)