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

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 $ep.$**

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?**

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.**

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$$\\$ $p_ A =2$ bar$\\$ $m _B =2g$$\\$ $p_ B =(3-2)=1$bar$\\$ (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?**

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 $27^ o C$?**

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 $27^ o C$?**

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 $STP$?**

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?**

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?**

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 $^{-1}$ ).**

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.**

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 $27^ o C. (R = 0.083\ \ bar\ \ dm ^3\ \ K ^{-1}\ \ mol ^{-1} ).$**

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 $R = 0.083\ \ bar\ dm ^{-3} K^ {-1} mol^{ -1}$ ).**

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$^{ -1}$**

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?**

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.**

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 $pV^ 2 T ^2 /n?$**

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.**

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?**

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.**

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.$\\$