Answer

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,

$ \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} bar \\ & =2.5 bar \end{aligned} $

Therefore, the minimum pressure required is 2.5 bar.

Answer

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} bar \\ & =0.8 bar \end{aligned} $

Therefore, the pressure would be 0.8 bar.

Answer

The equation of state is given by,

$p V=n R T$

Where,

p $\rightarrow$ Pressure of gas

V $\rightarrow$ Volume of gas

n $\rightarrow$ Number of moles of gas

R $\rightarrow$ Gas constant

T $\rightarrow$ Temerature of gas

From equation (i) we have,

$\frac{n}{V}=\frac{p}{R T}$

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

$\frac{m}{M V}=\frac{p}{R T}$

Where,

m $ \rightarrow$ mass of gas

M $\rightarrow$ Molar mass of gas

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

Thus, from equation (ii), we have $\frac{d}{M}=\frac{p}{R T}$

$\Rightarrow d=(\frac{M}{R T}) p$

Molar mass $(M)$ of a gas is always constant and therefore, at constant temperature

$(T), \frac{M}{R T}=$ constant.

$d=($ constant $) p$

$\Rightarrow d \propto p$

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

Answer

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

$d=\frac{M p}{R T}$

Now, density of oxide $(d_1)$ is given by,

$d_1=\frac{M_1 p_1}{R T}$

Where, $M_1$ and $p_1$ are the mass and pressure of the oxide respectively.

Density of dinitrogen gas $(d_2)$ is given by,

$d_2=\frac{M_2 p_2}{R T}$

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, $M_1=\frac{M_2 p_2}{p_1}$

$ \begin{aligned} & =\frac{28 \times 5}{2} \\ & =70 g / mol \end{aligned} $

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

Answer

For ideal gas $A$, the ideal gas equation is given by,

$p_A V=n_A R T$.

Where, $p_A$ and $n_A$ represent the pressure and number of moles of gas $A$.

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

$p_B V=n_B R T$

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

$$ p_A V=\frac{m_A}{M_A} R T \rightarrow \frac{p_A M_A}{m_A}=\frac{R T}{V} \ldots \ldots \tag{iii} $$

From equation (ii), we have

$$ p_B V=\frac{m_B}{M_B} R T \Rightarrow \frac{p_B M_B}{m_B}=\frac{R T}{V} \ldots \ldots \tag{iv} $$

Where, $M_A$ and $M_B$ are the molecular masses of gases $A$ and $B$ respectively.

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

$$ \frac{p_A M_A}{m_A}=\frac{p_B M_B}{m_B} \tag{v} $$

Given,

$ \begin{aligned} & m_A=1 g \\ & p_A=2 bar \\ & m_B=2 g \\ & p_B=(3-2)=1 bar \end{aligned} $

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

$ \begin{aligned} & \frac{2 \times M_A}{1}=\frac{1 \times M_B}{2} \\ & \Rightarrow 4 M_A=M_B \end{aligned} $

Thus, a relationship between the molecular masses of $A$ and $B$ is given by

$ 4 M_A=M_B . $

Answer

The reaction of aluminium with caustic soda can be represented as:

$2 Al+2 NaOH+2 H_2 O \longrightarrow 2 NaAlO_2+3 H_2$

$2 \times 27 g$ $3 \times 22400 mL$

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,

$p_1=1 atm$

$V_1=186.67 mL$

$T_1=273.15 K$

Let the volume of dihydrogen be ${ }^{V}$ at $p_2=0.987$ atm (since 1 bar $.=0.987 atm)$ and $T_2=20^{\circ} C=(273.15+20) K=$ 293.15 K.

Now,

$ \begin{aligned} \frac{p_1 V_1}{T_1} & =\frac{p_2 V_2}{T_2} \\ \Rightarrow 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} \\ & =202.98 mL \\ & =203 mL \end{aligned} $

Therefore, $203 mL$ of dihydrogen will be released.

Answer

It is known that,

$ p=\frac{m}{M} \frac{R T}{V} $

For methane $(CH_4)$,

$ \begin{aligned} p _{CH_4} & =\frac{3.2}{16} \times \frac{8.314 \times 300}{9 \times 10^{-3}} \begin{bmatrix} \text{ Since } 9 dm^{3}=9 \times 10^{-3} m^{3} \\ 27^{\circ} C=300 K \end{bmatrix} \\ & =5.543 \times 10^{4} Pa \end{aligned} $

For carbon dioxide $(CO_2)$,

$ \begin{aligned} p _{CO_2} & =\frac{4.4}{44} \times \frac{8.314 \times 300}{9 \times 10^{-3}} \\ & =2.771 \times 10^{4} Pa \end{aligned} $

Total pressure exerted by the mixture can be obtained as:

$ \begin{aligned} p & =p _{CH_4}+p _{CO_2} \\ & =(5.543 \times 10^{4}+2.771 \times 10^{4}) Pa \\ & =8.314 \times 10^{4} Pa \end{aligned} $

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

Answer

Let the partial pressure of $H_2$ in the vessel be $p _{H_2}$.

Now,

$ \begin{matrix} p_1=0.8 bar & p_2=p _{H_2}=? \\ V_1=0.5 L & V_2=1 L \end{matrix} $

It is known that,

$ \begin{aligned} 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} \\ & =0.4 bar \end{aligned} $

Now, let the partial pressure of $O_2$ in the vessel be $p _{O_2}$.

Now,

$p_1=0.7$ bar $\quad p_2=p _{O_2}=$ ?

$V_1=2.0 L \quad V_2=1 L$

$p 1 V 1=p 2$

Answer

Given,

$ \begin{aligned} & d_1=5.46 g / dm^{3} \\ & p_1=2 bar \\ & T_1=27^{\circ} C=(27+273) K=300 K \\ & p_2=1 bar \\ & T_2=273 K \\ & d_2=? \end{aligned} $

The density $(d_2)$ of the gas at STP can be calculated using the equation,

$ \begin{aligned} & d=\frac{M p}{R T} \\ & \begin{aligned} \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} \\ & =3 g dm^{-3} \end{aligned} \end{aligned} $

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

Answer

Given,

$p=0.1$ bar

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

$R=0.083$ bar $dm^{3} K^{{-1}} mol^{{-1}}$

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

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

$ \begin{aligned} p V & =n R T \\ \Rightarrow n & =\frac{p V}{R T} \\ & =\frac{0.1 \times 34.05 \times 10^{-3}}{0.083 \times 819} \\ & =5.01 \times 10^{-5} mol \end{aligned} $

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

Hence, the molar mass of phosphorus is $1247.5 g mol^{-1}$.

Answer

Let the volume of the round bottomed flask be $V$.

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

Now,

$V_1=V$

$T_1=27^{\circ} C=300 K$

$V_2=$ ?

$T_2=477^{\circ} C=750 K$

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{750 V}{300} $

$ =2.5 V $

Therefore, volume of air expelled out $=2.5 V - V=1.5 V$

Hence, fraction of air expelled out

$ =\frac{1.5 V}{2.5 V}=\frac{3}{5} $

Answer

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:

$ \begin{aligned} p V & =n R T \\ \Rightarrow T & =\frac{p V}{n R} \\ & =\frac{3.32 \times 5}{4 \times 0.083} \\ & =50 K \end{aligned} $

Hence, the required temperature is $50 K$.

Answer

Molar mass of dinitrogen $(N_2)=28 g mol^{{-1}}$

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 $=1.4 \times 3.01 \times 10^{23}$

$=4.214 \times 10^{23}$ electrons

Answer

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

Answer

Given,

Mass of dioxygen $(O_2)=8 g$

Thus, number of moles of

$ O_2=\frac{8}{32}=0.25 mole $

Mass of dihydrogen $(H_2)=4 g$

Thus, number of moles of

$ H_2=\frac{4}{2}=2 mole $

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

Given,

$V=1 dm^{3}$

$n=2.25 mol$

$dm^3 $ at $27^oC.R = 0.083 \text{ bar d}m^3K^{-1}mol^{-1}$

$T=27^{\circ} C=300 K$

Total pressure $(p)$ can be calculated as:

$p V=n R T$

$\Rightarrow p=\frac{n R T}{V}$

$ \begin{aligned} & =\frac{225 \times 0.083 \times 300}{1} \\ & =56.025 bar \end{aligned} $

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

Answer

Given,

Radius of the balloon, $r=10 m$

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

$=\frac{4}{3} \times \frac{22}{7} \times 10^{3}$

$=4190.5 m^{3}$ (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 $=4190.5 \times 1.2 kg$

$=5028.6 kg$

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

$m=\frac{M p V}{R T}$

Here,

$M=4 \times 10^{-3} kg mol^{-1}$

$p=1.66$ bar

$V=$ Volume of the balloon

$ =4190.5 m^{3} $

$R=0.083$ bar dm $^{3} K^{-1} mol^{-1}$

$T=27^{\circ} C=300 K$

Then, $m=\frac{4 \times 10^{-3} \times 1.66 \times 4190.5 \times 10^{3}}{0.083 \times 300}$

$ =1117.5 kg \text{ (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 $3811.1 kg$.

Answer

It is known that,

$ \begin{aligned} & p V=\frac{m}{M} R T \\ & \Rightarrow V=\frac{m R T}{M p} \end{aligned} $

Here,

$m=8.8 g$

$R=0.083$ bar $LK^{-1} mol^{{-1}}$

$T=31.1^{\circ} C=304.1 K$

$M=44 g$

$p=1$ bar

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

$ \begin{aligned} & =5.04806 L \\ & =5.05 L \end{aligned} $

Hence, the volume occupied is $5.05 L$.

Answer

Volume ( $V$ ) occupied by dihydrogen is given by,

$ \begin{aligned} V & =\frac{m}{M} \frac{R T}{p} \\ & =\frac{0.184}{2} \times \frac{R \times 290}{p} \end{aligned} $

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

$ \begin{aligned} V & =\frac{m}{M} \frac{R T}{p} \\ & =\frac{2.9}{M} \times \frac{R \times 368}{p} \end{aligned} $

According to the question,

$ \begin{aligned} & \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} \\ & \Rightarrow M=\frac{2.9 \times 368 \times 2}{0.184 \times 290} \\ & =40 g mol^{-1} \end{aligned} $

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

Answer

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 \text{ moles } \text{ and the number of moles of } $

dioxygen,

$ n _{O_2}=\frac{80}{32}=2.5 moles $

Given,

Total pressure of the mixture, $p _{\text{total }}=1$ bar

Then, partial pressure of dihydrogen,

$ \begin{aligned} p _{H_2} & =\frac{n _{H_2}}{n _{H_2}+n _{O_2}} \times P _{\text{total }} \\ & =\frac{10}{10+2.5} \times 1 \\ & =0.8 bar \end{aligned} $

Hence, the partial pressure of dihydrogen is 0.8 bar .

Answer

The $SI$ unit 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 $\frac{p V^{2} T^{2}}{n}$ is given by,

$=\frac{(Nm^{-2})(m^{3})^{2}(K)^{2}}{mol}$

$=Nm^{4} K^{2} mol^{-1}$

Answer

Charles’ 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 ${ }^{\circ} C$ ) is a straight line. If this line is extended to zero volume, then it intersects the temperature-axis at $-273^{\circ} C$. In other words, the volume of any gas at $-273^{\circ} C$ is zero. This is because all gases get liquefied before reaching a temperature of $-273^{\circ} C$. Hence, it can be concluded that $-273^{\circ} C$ is the lowest possible temperature.

Answer

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 $CO_2$.

Answer

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.