 # 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. \\ AtSTP , \\ 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} 11 A student forgot to add the reaction mixture to the round bottomed flask at 27^ o 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 ^o C. What fraction of air would have been expelled out? ##### Solution : 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,$\\$ $\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\ = V_2=\dfrac{V_1T_2}{T_1}\\ =\dfrac{750V}{300}\\ =2.5V$$\\ Therefore, volume of air expelled out = 2.5 V – V = 1.5 V$$\\$ Hence, fraction of air expelled out=$\dfrac{1.5V}{2.5V}=\dfrac{3}{5}$

12   Calculate the temperature of 4.0 mol of gas occupying 5 dm $^3$ at 3.32 bar. (R = 0.083 bar dm$^ 3$ K$^{ -1}$ mol $^{-1}$ ).

##### Solution :

Given,$\\$ $n = 4.0 mol\\ V = 5 dm^ 3\\ p = 3.32 bar\\ R = 0.083 bar\ \ dm^ 3\ \ K^{ -1}\ \ mol ^{-1}$ The temperature (T) can be calculated using the ideal gas equation as:$\\$ $pV = nRT\\ = T=\dfrac{pV}{nR}\\ =\dfrac{3.32*5}{4*0.083}\\ =50K$ $\\$ Hence, the required temperature is$50 K.$

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

##### Solution :

Molar mass of dinitrogen $(N _2 ) = 28 g mol^{ -1}$$\\ Thus, 1.4 g\ \ of \ \ N _2 =\dfrac{1.4}{28}=0.05 mol\\ = 0.05 * 6.02 * 10 ^{23} number of molecules\\ = 3.01 * 10^{ 23} number of molecules\\ Now, 1 molecule of N _2 contains 14 electrons.\\ Therefore, 3.01 * 10^{ 23} molecules of N _2 contains = 14 * 3.01 * 10^{23}$$\\$ $= 4.214 * 10^{ 23}$ electrons$\\$

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

Avogadro number = $6.02 * 10^{ 23}$$\\ Thus, time required\\ =\dfrac{6.02*10^{23}}{10^{10}}s\\ =6.02×10^{ 23} s\\ =\dfrac{6.02×10^{ 23}}{60×60×24×365} years \\ =1.909×10 ^6 years Hence, the time taken would be =1.909×10^ 6 years . 15 Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm^ 3 at 27^ o C. (R = 0.083\ \ bar\ \ dm ^3\ \ K ^{-1}\ \ mol ^{-1} ). ##### Solution : Given,\\ Mass of dioxygen (O_ 2 ) = 8 _g$$\\$ Thus, number of moles of $O _2=\dfrac{8}{32}=0.25 \ mole$ $\\$ Mass of dihydrogen $(H _2 ) = 4 _g\\ H_2=\dfrac{4}{2}=2\ mole$ $\\$ Therefore, total number of moles in the mixture =$0.25 + 2 2.25$ mole$\\$ Given,$\\$ $V = 1 dm^3\\ n = 2.25 \ mol\\ R = 0.083\ bar\ dm^ 3\ K^{ -1}\ mol^{ -1}\\ T = 27^ o C = 300 K$$\\ Total pressure (p) can be calculated as:\\ pV= nRT\\ =p=\dfrac{nRT}{V}\\ =\dfrac{225 * 0.083 * 300}{1}\\ = 56.025 bar$$\\$ Hence, the total pressure of the mixture is 56.025 bar.

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 27$^ o$ C. (Density of air = $1.2 kg\ m^{ -3}$ . And $R = 0.083\ \ bar\ dm ^{-3} K^ {-1} mol^{ -1}$ ).

##### Solution :

Given,$\\$ Radius of the balloon, r = 10 m$\\$ $\therefore$Volume of the balloon =$\dfrac{4}{3}\pi r^3\\ =\dfrac{4}{3}*\dfrac{22}{7}*10^{23}\\ =4190.5m^3$(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 * 1.2 kg$$\\ = 5028.6 kg$$\\$ Now, mass of helium (m) inside the balloon is given by,$\\$ $m=\dfrac{MpV}{RT}$ $\\$ $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^ o C = 300 K$$\\$ Then,$m=\dfrac{4×10^{ -3} ×1.66×4190.5×10^ 3}{0.083×300}\\ =1117.5kg$ (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.$\\$

17   Calculate the volume occupied by 8.8 g of $CO _2$ at $31.1^ o C$ and 1 bar pressure. $R = 0.083\ bar\ L K^{ -1 }$mol$^{ -1}$

##### Solution :

It is known that,$\\$ $pV=\dfrac{m}{N}RT\\ =V=\dfrac{mRT}{Mp}$ $\\$ Here,$\\$ $m = 8.8 g\\ R = 0.083\ bar\ LK^{ -1} mol^{ -1}\\ T = 31.1^ o C = 304.1 K\\ M = 44 g\\ p = 1 bar$ $\\$ Thus, Volume (V) =$\dfrac{8.8 * 0.083 * 304.1}{44 * 1}\\ = 5.04806 L\\ = 5.05 L\\$ Hence, the volume occupied is$5.05 L.$

18   $2.9 g$ of gas at $95^ o C$ occupied the same volume as $0.184 g$ of dihydrogen at $17 ^o C$, at the same pressure. What is the molar mass of the gas?

##### Solution :

Volume (V) occupied by dihydrogen is given by,$\\$ $V=\dfrac{m}{M}\dfrac{RT}{p}\\ =\dfrac{0.184}{2}*\dfrac{R*290}{p}$ $\\$ Let M be the molar mass of the unknown gas. Volume (V) occupied by the unknown gas can be calculated as:$\\$ $V=\dfrac{m}{M}\dfrac{RT}{p}\\ =\dfrac{2.9}{M}*\dfrac{R*368}{p}$ $\\$ According to the equation,$\\$ $\dfrac{0.184}{2}*\dfrac{R*290}{p}=\dfrac{2.9}{M}*\dfrac{R*368}{P}\\ =\dfrac{0.184*290}{2}=\dfrac{2.9*368}{M}\\ =M=\dfrac{2.9*368*2}{0.184*290}\\ =40g \ mol^{-1}$ $\\$ Hence, the molar mass of the gas is $40 g\ mol^{ -1}$ .

19   A mixture of dihydrogen and dioxygen atone bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

##### Solution :

Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80 g.$\\$ Then, the number of moles of dihydrogen, $n_{H_2}=\dfrac{20}{2}=10 \ moles$and the number of moles of dioxygen,$n_{O_2}=\dfrac{80}{32}=2.5 \ moles.$ $\\$ Given,$\\$ Total pressure of the mixture, $P_{ total }= 1$ bar$\\$ Then, partial pressure of dihydrogen,$\\$ $P_{H_2}=\dfrac{n_{H_2}}{n_{H_2}+n_{O_2}}*P_{total}\\ =\dfrac{10}{10+2.5}*1\\ =0.8\ bar$$\\$ Hence, the partial pressure of dihydrogen is 0.8 bar.

20   What would be the SI units for the quantity $pV^ 2 T ^2 /n?$

##### Solution :

The SI units 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. Therefore, the SI unit for quantity$\dfrac{pV^2T^2}{n}$ is given by,$\\$ $=\dfrac{(Nm^{-2})(m^3)^2(K)^2}{mol}\\ =Nm^4K^2 mol^{-1}$

21   In terms of Charles’ law explain why $-273^ o C$ is the lowest possible temperature.

##### Solution :

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

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.

22   Critical temperature for carbon dioxide and methane are $31.1^ o C$ and $– 81.9^ o C$ respectively.Which of these has stronger intermolecular forces and why?

##### Solution :

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 $CO _2$ .

23   Explain the physical significance of Van der Waals parameters.

##### Solution :

The vander 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( Vander waals force) The equation is

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.$\\$