 # States of Matter

## Chemistry Class 11

### NCERT

1   What will be the minimum pressure required to compress $500 dm ^3$ of air at $1$ bar to $200 dm ^3$ at $30^ o C$?

##### Solution :

Given,$\\$ Initial pressure, $p _1 = 1$bar$\\$ Initial volume, $V _1 = 500 dm ^3$$\\ Final volume, V _2 = 200 dm^3$$\\$ Since the temperature remains constant, the final pressure $(p^ 2 )$ can be calculated using Boyle’s law.$\\$ According to Boyle’s law,$\\$ $p_1V_1=p_2V_2$$\\ \Rightarrow p_2=\dfrac{p_1V_1}{V_2}$$\\$ $=\dfrac{1*500}{200}$bar$\\$ $=2.5$ bar$\\$ Therefore, the minimum pressure required is $2.5$ bar.

2   A vessel of $120 mL$ capacity contains a certain amount of gas at $35^o C$ and $1.2$ bar pressure. The gas is transferred to answer vessel of volume $180 mL$ at $35^o C.$ What would be its pressure?

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$$\\ p_2=\dfrac{p_1V_1}{V_2}$$\\$ $=\dfrac{1.2*120}{180}bar$$\\ =0.8 bar\\ Therefore, the pressure would be 0.8 bar. 3 Using the equation of state = pV = nRT; show that at a given temperature density of a gas is proportional to gas pressure ep. ##### Solution : The equation of state is given by,\\ pV = nRT ............ (i)$$\\$ Where,$\\$ p $\to$ Pressure of gas$\\$ V $\to$ Volume of gas$\\$ n $\to$ Number of moles of gas$\\$ R $\to$Gas constant $\\$ T $\to$ Temperature of gas$\\$ From equation (i) we have,$\\$ $\dfrac{n}{V}=\dfrac{p}{RT}$$\\ Replacing n with \dfrac{m}{M},we have\\ \dfrac{m}{MV}=\dfrac{p}{RT}....(ii)$$\\$ Where,$\\$ m$\to$ Mass of gas$\\$ M$\to$Molar mass of gas$\\$ But,$\dfrac{m}{V}=d(d=$\text{density of gas})$\\$ Thus, from equation(ii),we have$\\$ $\dfrac{d}{M}=\dfrac{p}{RT}$$\\ \Rightarrow d=(\dfrac{M}{RT})p$$\\$ Molar mass (M) of gas is always constant and therefore, at constant temperature(T),$\dfrac{M}{RT}=$ constant,$\\$ d=(constant)p$\\$ $\Rightarrow d \alpha p$$\\ Hence, at a given temperature, the density (d) of gas is proportional to its pressure (p) 4 At 0^o C, the density of certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide? ##### Solution : Density (d) of substance at temperature (T) can be given by the expression,\\ d=\dfrac{Mp}{RT} Now, density of oxide (d _1 ) is given by,\\ d_1=\dfrac{M_1p_1}{RT}$$\\$ 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=\dfrac{M_2p_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_1p_1=M_2p_2$$\\ Given,\\ p_1=2 bar\\ p_2=5 bar \\ Molecular mass of nitrogen, M _2 = 28 g/mol$$\\$ Now $,M_1=\dfrac{M_2P_2}{P_1}$$\\ =\dfrac{28*5}{2}$$\\$ $=70g/mol$$\\ Hence, the molecular mass of the oxide is 70 g/mol. 5 Pressure of 1 g of an ideal gas A\ 27 ^o C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses. ##### Solution : For ideal gas A, the ideal gas equation is given by,\\ p_B V = n _B RT ......... (ii)$$\\$ 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=\dfrac{m_A}{M_b}RT $$\\ \Rightarrow\dfrac{p_AM_A}{m_A}=\dfrac{RT}{V}......(iii) \\ From equation (ii),we have\\ p_BV=\dfrac{m_B}{M_B}RT$$\\$ $\Rightarrow \dfrac{p_BM_B}{m_B}=\dfrac{RT}{V}.....(iv)$ $\\$ Where, $M _A$ and $M _B$ are the molecular masses of gases $A$ and $B$ respectively.$\\$ Now, from equations (iii) and (iv), we have$\\$ $\dfrac{p_AM_B}{m_B}=\dfrac{p_BM_B}{m_B}........(v)$$\\ Given,\\ m _A =1g$$\\$ $p_ A =2$ bar$\\$ $m _B =2g$$\\ p_ B =(3-2)=1bar\\ (Since total pressure is 3 bar)\\ Substituting these values in equation (v), we have\\ \dfrac{2*M_A}{1}=\dfrac{1*M_B}{1}$$\\$ $\Rightarrow 4M_A=M_B$$\\$ Thus, a relationship between the molecular masses of $A$ and $B$ is given by $4 M _A = M _B$

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

##### Solution :

The reaction of aluminum with caustic soda can be represented as:$\\$ $2Al+2NaOH+2H_ 2 0 \to 2NaAlO _2 +3H_ 2$ $\\$ $2×27g \qquad \qquad \qquad \qquad 3×22400mL$ $\\$ At $STP (273.15 K$ and $1$ atm$), 54 g (2 *27 g)$ of $Al$ gives $3 *22400 mL$ of $H_ 2$.$\\$ $\therefore 0.15g \ Al$ gives $\dfrac{3*22400*0.15}{54}mL$ of $H_2$ i.e.,$186.67 mL$ of $H_2.$ $\\$ At$STP ,$ $\\$ $p _1 =1$ atm $\\$ $V _1 =186.67 mL$ $\\$ $T_ 1 =273.15K$ $\\$ Let the volume of dihydrogen be $V_ 2$ at $p _2 = 0.987$ atm (since $1$ bar = $0.987$ atm) and $T _2 = 20 ^o C$ $\\$ $= (273.15 + 20) K = 293.15 K.$ $\\$ $\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$ $\\$ $PV_2=\dfrac{P_1V_1T_2}{P_2T_1}$ $\\$ $=\dfrac{1*186.67*293.15}{0.987*273.15}$ $\\$ $=202.98 mL$ $\\$ $=203 mL$ $\\$ Therefore, $203 mL$ of dihydrogen will be released.

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 $9 d^ 3$ flask at $27^ o C$?

##### Solution :

It is known that,$\\$ $p=\dfrac{m}{M}\dfrac{RT}{V}$ $\\$ For methane $(CH_ 4 ),$ $\\$ $P_{CH_4}=\dfrac{3.2}{16}*\dfrac{8.314*300}{9*10^{-3}}$[Since $9dm =9×10^{-3} m^3$]$\\$ $=5.543*10^4 Pa$ $\\$ For carbon dioxide $(CO_ 2 ),$ $\\$ $P_{CO_4}=\dfrac{4.4}{44}*\dfrac{8.314*300}{9*10^{-3}}$ $\\$ $=2.771*10^4 Pa$ $\\$ The pressure exerted by the mixture can be obtained as:$\\$ $P=P_{CH_4}+P_{CO_2}$ $\\$ $=(5.543*10^4+2.771*10^4)Pa$ $\\$ $=8.314*10^4 Pa$ $\\$ Hence, the total pressure exerted by the mixture is $=8.314 * 10^ 4 Pa .$

8   What will be the pressure of he gaseous mixture when $0.5 L$ of $H _2$at $0.8$ bar and $2.0 L$ of dioxygen at $0.7$ bar are introduced in a $1L$ vessel at $27^ o C$?

##### Solution :

Let the partial pressure of $H_ 2$ in the vessel be $p_ {H _2} .$ $\\$ Now,$\\$ $p_ 1 = 0.8$ bar$\\$ $p_ 2 = p_{ H _2 }=?$ $\\$ $V_ 1 = 0.5L$ $\\$ $V_ 2 =1L$ $\\$ It is known that,$\\$ $p_ 1 V _1 = p_ 2 V_ 2$ $\\$ $\Rightarrow p_2=\dfrac{p_1V_1}{V_2}$ $\\$ $\Rightarrow p_{H_2}=\dfrac{0.8*0.5}{1}$ $\\$ $=0.4$ bar $\\$ Now, let the partial pressure of $O_ 2$ in the vessel be $p_ {O_ 2} .$ $\\$ $p_ 1 =0.7$ bar $\\$ $p_2=p_{o_2}=?$ $\\$ $V_1=2.0L$ $\\$ $V_2=1L$ $\\$ $p_1V_1=p_2V_2$ $\\$ $\Rightarrow p_2=\dfrac{p_1V_1}{V_2}$ $\\$ $\Rightarrow p_{o_2}=\dfrac{0.7*20}{1}$ $\\$ $=0.4$ bar $\\$ Total pressure of the gas mixture in the vessel can be obtained as:$\\$ $p_{total}=p_{H_2}+p_{o_2}$ $\\$ $=0.4+1.4$ $\\$ $=1.8$ bar $\\$ Hence, the total pressure of the gaseous mixture in the vessel is $1.8$ bar.

9   Density of a gas is found to be $5.46 g/dm ^3$ at $27 ^o C$ at $2$ bar pressure. What will be its density at $STP$?

##### Solution :

Given,$\\$ $d_ 1 = 5.46 g / dm ^3$ $\\$ $p _1 = 2$ bar $\\$ $T _1 = 27 ^o C = (27 + 273) K = 300 K$ $\\$ $p_ 2 = 1$ bar$\\$ $T_ 2 = 273 K$ $\\$ $d _2 = ?$ $\\$ The density ($d_ 2$ ) of the gas at $STP$ can be calculated using the equation,$\\$ $d=\dfrac{Mp}{RT}$ $\\$ $\therefore \dfrac{d_1}{d_2}=\dfrac{\dfrac{Mp_1}{RT_1}}{\dfrac{Mp_2}{RT_2}}$ $\\$ $\Rightarrow \dfrac{d_1}{d_2}=\dfrac{p_1T_2}{p_2T_1}$ $\\$ $\Rightarrow d_2=\dfrac{p_2T_1d_1}{p_1T_2}$ $\\$ $=\dfrac{1*300*5.46}{2*273}$ $\\$ $=3 gdm^{-3}$ $\\$ Hence, the density of the gas at $STP$ will be $3 g dm ^{-3 }$.

10   $34.05 mL$ of phosphorus vapour weighs $0.0625 g$ at $546 ^o C$ and $0.1$ bar pressure. What is the molar mass of phosphorus?

##### Solution :

Given,$\\$ $p = 0.1$ bar $\\$ $V = 34.05 mL = 34.05 * 10 ^{-3} L = 34.05 * 10^{ -3} dm^{ -3}$ $\\$ $R =0.083 bar \ \ dm ^3 K ^{-1} \ \ mol^{ -1}$ $T = 546 ^o C = (546 + 273) K = 819 K$ $\\$ The number of mass (n) can be calculated using the ideal gas equation as:$\\$ $pV = nRT$ $\\$ $\Rightarrow n=\dfrac{pV}{RT}=\dfrac{0.1*34.05*10^{-3}}{0.083*819}$ $\\$ $=5.01*10^{-5}$ mol$\\$ Therefore, molar mass of phosphorus =$\dfrac{0.0625}{5.01*10^{-5}}=1247.5 g \ mol^{-1}$ $\\$ Hence, the molar mass of phosphorus is $1247.5 g\ mol ^{- 1}$