Gaseous State

Let the molar mass of He be $M$.

Then, according to the question, the molar mass of Ar is

$ 10M $

Since both gases behave ideally, for a mixture we can use

$ PV = nRT $

where $n$ is the total number of moles in the cylinder.

At the same temperature, if the volumes of the two cylinders are equal, then from

$ PV = nRT $

we get

$ P \propto n $

because $V$ and $T$ are same.

Now, it is given that the pressure on the first cylinder is 5 times that on the second:

$ P_1 = 5P_2 $

Hence,

$ n_1 = 5n_2 $

Now calculate total moles in each cylinder.

For the first cylinder:

• mass of He $= m_1$, so moles of He $= \dfrac{m_1}{M}$

• mass of Ar $= m_2$, so moles of Ar $= \dfrac{m_2}{10M}$

Thus,

$ n_1 = \frac{m_1}{M} + \frac{m_2}{10M} $

For the second cylinder:

• mass of He $= m_2$, so moles of He $= \dfrac{m_2}{M}$

• mass of Ar $= m_1$, so moles of Ar $= \dfrac{m_1}{10M}$

Thus,

$ n_2 = \frac{m_2}{M} + \frac{m_1}{10M} $

Using $n_1 = 5n_2$:

$ \frac{m_1}{M} + \frac{m_2}{10M} = 5\left(\frac{m_2}{M} + \frac{m_1}{10M}\right) $

Multiply throughout by $10M$:

$ 10m_1 + m_2 = 50m_2 + 5m_1 $

$ 10m_1 - 5m_1 = 50m_2 - m_2 $

$ 5m_1 = 49m_2 $

Therefore,

$ \frac{m_1}{m_2} = \frac{49}{5} $

So, the required value is

$ \boxed{\frac{49}{5}} $

Molar volume $\rightarrow V_m$ (volume occupied by I mole of the gas at the given temperature and pressure). $a=6.0 \mathrm{dm}^6 \mathrm{~atm} \mathrm{~mol}^{-2}$ $ b=0.060 \mathrm{dm}^3 \mathrm{~mol}^{-1} $ Temperature, $T=300 \mathrm{~K}$ Pressure, $P=300 \mathrm{~atm}$ $ R=0.082 \mathrm{~d} \mathrm{~m}^3 \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} $ $ \frac{V_m^2(\text { coefficient })}{V_m(\text { coefficient })}=?\left(\mathrm{~mol} \mathrm{dm}^{-3}\right) $ Van der Walls equation is $ \begin{aligned} & \left(P+\frac{a n^2}{V^2}\right)(V-n b)=n k T \\ & n=1 \mathrm{~mol} \quad, V=V_m \end{aligned} $ So, $\left(P+\frac{a}{V_m^2}\right)\left(V_m-b\right)=R T$The equation can be expanded as $ \begin{aligned} & P V_m-P b+\frac{a}{V_m^2} \times V_m-\frac{a}{V_m^2}=R T \\ & P V_{1 m}-P b+\frac{a}{V_m}+\frac{a b}{V_m^2}=R T \end{aligned} $ Multiply by $\mathrm{Vm}^2$$ \begin{aligned} & P V_m \times V_m^2-P b \times V_m^2+\frac{a}{V_m} \times V_m^2-\frac{a b}{V_m^2} \times V_m^2=R T \times V_m^2 \\ & P V_m^3-P b V_m^2+a V_m-a b=R T V_m^2 \\ & P V_m^3-P b V_m^2+a V_m-a b-R T V_m^2=0 \\ & P V_m^3-(P b+R T) V_m^2+a V_m-a b=0 \end{aligned} $Coefficient of $V_m^2 \rightarrow-(P b+R T)$ coefficient of $v_m \longrightarrow a$ The ratio, $ \begin{aligned} & \frac{\text { Coefficient of } V_m^2}{\text { coefficient of } V_m} \\ & =\frac{-(P b+R T)}{a} \end{aligned} $ $ \text { Substitute the values as } $$ \frac{\text { Coefficient of } \mathrm{Vm}^2}{\text { Coefficient of } \mathrm{Vm}_{\mathrm{m}}}\begin{aligned} & -\left(300 \mathrm{~atm} \times 0.060 \mathrm{dm}^3 \mathrm{~mol}^{-1}+\right. \\ = & \frac{0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol} \mathrm{k}^{-1} \times 300K )\mathrm{}}{6.0 \mathrm{dm}^6 \mathrm{~atm} \mathrm{~mol}{ }^{-2}} \end{aligned} $$ \frac{\text { Coesficient of } V_m^2}{\text { coefficient of } V_m}=\frac{-18 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~mol}^{-1}+24.6 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol}^{-1}}{6.0 \mathrm{dm} \mathrm{dmol}^6 \mathrm{~atm} \mathrm{~mol}^2} $$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\frac{42.6 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol}}{-6.0 \mathrm{dm}^6 \mathrm{~atm} \mathrm{~mol}^{-2}} $$ \begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,& =-7.1 \mathrm{dm}^3 \mathrm{~mol}^{-1} \mathrm{dm}^{-6} \mathrm{~mol}^2 \\ & =-7.1 \mathrm{dm}^{3-6} \mathrm{~mol}^{-1+2} \\ & =-7.1 \mathrm{dm}^{-3} \mathrm{~mol}^1 \\ & =-7.1 \mathrm{~mol} \mathrm{dm}^{-3} \end{aligned} $The ratio is -7.1 Answer: $\quad-7.1$

$ \begin{aligned} & \mathrm{z}=\frac{\mathrm{PV}}{\mathrm{nRT}} ; \mathrm{n}=\frac{\mathrm{PV}}{\mathrm{ZRT}} \\\\ & \mathrm{Z}_1=1.07, \mathrm{P}_1=100 \mathrm{~atm}, \mathrm{~V}_1=0.15 \mathrm{~L}, \mathrm{~T}_1=500 \mathrm{~K} \\\\ & \mathrm{Z}_2=1.4, \mathrm{P}_2=300 \mathrm{~atm}, \mathrm{~T}_2=300 \mathrm{~K}, \mathrm{~V}_2=? \\\\ & \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{Z}_1 \mathrm{RT}_{\mathrm{I}}}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{Z}_2 \mathrm{RT}_2}=\mathrm{n} \\\\ & \mathrm{V}_2=\frac{1.4}{1.07} \times .03 \\\\ &=392 \times 10^{-4} \mathrm{dm}^3 \end{aligned} $

The root mean square speed ($v_{rms}$) of a gas can be calculated using the following formula:

$v_{rms} = \sqrt{\frac{3kT}{M}}$

where:

• $k$ is Boltzmann's constant ($1.38 \times 10^{-23}$ J/K),
• $T$ is the temperature in Kelvin,
• $M$ is the molar mass of the gas in kilograms per mole.

The most probable speed ($v_{mp}$) of a gas can be calculated using the following formula:

$v_{mp} = \sqrt{\frac{2kT}{M}}$

Given that the root mean square speed of gas X at 600 K is equal to the most probable speed of gas Y at 90 K, we can set the two equations equal to each other and solve for the molar mass of gas Y:

$\sqrt{\frac{3kT_X}{M_X}} = \sqrt{\frac{2kT_Y}{M_Y}}$

Square both sides to eliminate the square root:

$\frac{3kT_X}{M_X} = \frac{2kT_Y}{M_Y}$

We are asked to find $M_Y$, so let's rearrange the equation to solve for $M_Y$:

$M_Y = \frac{2kT_YM_X}{3kT_X}$

We know that $T_X = 600$ K, $M_X = 40$ g/mol $= 40 \times 10^{-3}$ kg/mol (converted from grams to kilograms), $T_Y = 90$ K, and $k = 1.38 \times 10^{-23}$ J/K. Substituting these values into the equation gives:

$M_Y = \frac{2 \times 1.38 \times 10^{-23} \text{J/K} \times 90 \text{K} \times 40 \times 10^{-3} \text{kg/mol}}{3 \times 1.38 \times 10^{-23} \text{J/K} \times 600 \text{K}}$

Solving this equation gives:

$M_Y = 0.004 \text{kg/mol}$

To convert this value back into grams per mole, multiply by 1000:

$M_Y = 4 \text{g/mol}$

So, the molar mass of gas Y is approximately 4 g/mol.

To find the pressure at which the volume decreases by 40%, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

Let's assume the initial volume of the gas is V. According to Boyle's Law:

P₁V₁ = P₂V₂

Where P₁ is the initial pressure (940.3 mm Hg), V₁ is the initial volume, P₂ is the final pressure (which we need to find), and V₂ is the final volume (which is 40% less than the initial volume, or 0.6V).

Plugging in the values, we have:

940.3 mm Hg × V₁ = P₂ × (0.6V)

Now we can solve for P₂:

P₂ = (940.3 mm Hg × V₁) / (0.6V)

Since the question does not provide the initial volume (V₁), we cannot calculate the exact pressure. However, we can determine the pressure ratio when the volume decreases by 40%.

P₂ = (940.3 mm Hg × V₁) / (0.6V)

We can simplify this by canceling out V:

P₂ = 940.3 mm Hg / 0.6

P₂ ≈ 1567 mm Hg (nearest integer)

Therefore, the pressure at which the volume decreases by 40% is approximately 1567 mm Hg.

Dalton's law of partial pressure states that whenever two or more gases, which do not react chemically, are enclosed in vessel, the total pressure is equal to sum of partial pressure of each gas.

From Dalton's partial pressure law,

$ \begin{aligned} & P_f V_f=P_1 V_1+P_2 V_2+P_3 V_3 \\\\ & P_f \times 9=2 \times 2+4 \times 3+3 \times 4 \\\\ & P_f=\frac{28}{9}=3.11=3 \end{aligned} $

$ \begin{aligned} & P_{\text {Total }}=740 \mathrm{~mm} \text { of } \mathrm{Hg} \\\\ & \mathrm{P}_{\mathrm{X}}= (\text { mole fraction of } X) \mathrm{P}_{\text {Total }} \\\\ & \mathrm{n}_X=\frac{0.6}{20}=0.03 \\\\ & \mathrm{n}_Y=\frac{0.45}{45}=0.01 \\\\ & \text { Mole fraction of } X=\frac{0.03}{0.01+0.03}=\frac{3}{4} \\\\ & \text { Partial pressure of } X=\frac{3}{4} \times 740 \\\\ & =555 \mathrm{~mm} \text { of } \mathrm{Hg} \end{aligned} $

Moles = 0.3 $\times$ 0.2

Volume at STP = $0.3\times0.2\times22.7$

= 1.362 litre

= 1362 mL

At
(a) $\rightarrow \mathrm{CO}_2$ exist as gas

(b) $\rightarrow$ liquefaction of $\mathrm{CO}_2$ starts

(c) $\rightarrow$ liquefaction ends

(d) $\rightarrow \mathrm{CO}_2$ exist as liquid.

Between (b) & (c) $\rightarrow$ liquid and gaseous $\mathrm{CO}_2$ co-exist. As volume changes from (b) to (c) gas decreases and liquid increases.

(A), (C) $\rightarrow$ correct

For gas : $\mathrm{Z}=0.5, \mathrm{~V}_{\mathrm{m}}=0.4 \mathrm{~L} / \mathrm{mol}$

$ \begin{aligned} & \mathrm{T}=800 \mathrm{~K}, \mathrm{P}=\mathrm{X} \text { atm. } \\\\ \Rightarrow & \mathrm{Z}=\frac{\mathrm{PV}_{\mathrm{m}}}{\mathrm{RT}} \\\\ \Rightarrow & \frac{\mathrm{X}(0.4)}{0.08 \times 800}=0.5 \\\\ \Rightarrow & \mathrm{X}=80 \end{aligned} $

For ideal gas, $\mathrm{PV}_{\mathrm{m}}=\mathrm{RT}$

$ \Rightarrow \mathrm{V}_{\mathrm{m}}=\frac{\mathrm{RT}}{\mathrm{P}}=\frac{0.08 \times 800}{80}=0.8 \mathrm{~L} \mathrm{~mol}^{-1}=\mathrm{y} $

Then, $\frac{x}{y}=\frac{80}{0.8}=100$.

$P_{\mathrm{Ne}}=\left[P_{\text {Total }}\right] X_{\mathrm{Ne}}$

$20=25\left[X_{N e}\right]$

${\left[X_{N e}\right]=\frac{4}{5} }$

$\Rightarrow\left[\frac{\frac{200}{20}}{\frac{200}{20}+\frac{x}{32}}\right]=\frac{4}{5}$

$ \Rightarrow $$\frac{10}{10+\left(\frac{x}{32}\right)}=\frac{4}{5}$

$ \Rightarrow $$50=40+\frac{x}{8}$

$ \Rightarrow $$400=320+x$

$ \Rightarrow $$x=80$ gram

For 1 mole at high pressure

$P(V-b)=R T$

$\mathrm{PV}-\mathrm{Pb}=\mathrm{RT}$

$\frac{\mathrm{PV}}{\mathrm{RT}}=1+\frac{\mathrm{Pb}}{\mathrm{RT}}$

$Z=1+\frac{\mathrm{Pb}}{\mathrm{RT}}$

$1=\frac{99(\mathrm{~b})}{0.083 \times 298}$

$b=\frac{0.083 \times 298}{99} \simeq 0.249 \simeq 25 \times 10^{-2}$

Number of moles of mixture of $\mathrm{H}_{2}$ and $\mathrm{He}$

$ \begin{aligned} &=\frac{P V}{R T} \\ &=\frac{6 \times 10^{5} \times 0.0125}{8.3 \times 300}=3 \end{aligned} $

Let the mass of He in $10 \mathrm{~g}$ mixture be $\mathrm{xg}$

$\therefore \frac{x}{4}+\frac{10-x}{2}=3$

On solving $x=8 \mathrm{~g}$

$\therefore$ Mass of $\mathrm{He}$ in the mixture $=8 \mathrm{~g}$

$40 \% \,\mathrm{w} / \mathrm{w}$ hydrogen gas is given in mixture of $\mathrm{H}_{2}$ and oxygen.

Wt. of $\mathrm{H}_{2}=40 \mathrm{~g}$

Wt. of $\mathrm{O}_{2}=60 \mathrm{~g}$

$\chi_{\mathrm{H}_{2}}=\frac{\mathrm{n}_{\mathrm{H}_{2}}}{\mathrm{n}_{\mathrm{H}_{2}}+\mathrm{n}_{\mathrm{O}_{2}}}$

$=\frac{\frac{40}{2}}{\frac{40}{2}+\frac{60}{32}}$

$=\frac{20}{20+1.875}$

$=\frac{20}{21.875}=0.914$

$\mathrm{P}_{\mathrm{H}_{2}}=\chi_{\mathrm{H}_{2}} \times \mathrm{P}_{\mathrm{T}}$

$=0.914 \times 2.2$

$=2.01 \simeq 2 \mathrm{bar}$

From ideal gas equation,

$\mathrm{PV}=\mathrm{nRT}$

$P=2 \times 10^{6} \mathrm{~Pa}$

$V=2 \mathrm{dm}^{3}=2 \times 10^{-3} \mathrm{~m}^{3}$

$\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

$\mathrm{n}=\frac{11}{44} \mathrm{~mol}$

$2 \times 10^{6} \times 2 \times 10^{-3}=\frac{11}{44} \times 8.3 \times \mathrm{T}$

$\mathrm{T}=1927.7 \mathrm{~K}$

$\mathrm{T}\left(\right.$ in $\left.^{\circ} \mathrm{C}\right)=1927.7-273 \simeq 1655^{\circ} \mathrm{C}$

To solve this problem, we need to determine the new pressure of the moist gas when the volume of the container is doubled while maintaining the same temperature. First, we need to separate the contributions of the gas and the water vapor to the initial pressure and then apply Boyle's Law for the gas component only. The steps are as follows:

1. Initial pressure of the moist gas (gas + water vapor) is given as $4 \, \text{atm}$.

The vapour pressure of water at $27^{\circ} \mathrm{C}$ is $0.4 \, \text{atm}$.

2. The partial pressure of the dry gas component can be calculated by subtracting the vapor pressure of water from the initial pressure:

$ P_{\text{dry gas (initial)}} = 4 \, \text{atm} - 0.4 \, \text{atm} = 3.6 \, \text{atm} $

3. When the volume of the container is doubled, the pressure of the dry gas component will change according to Boyle's Law, which states:

$ P_1 V_1 = P_2 V_2 $

Since the temperature remains constant and the volume is doubled (i.e., $V_2 = 2V_1$), we can find the new pressure of the dry gas ($P_{\text{dry gas (new)}}$) with the equation:

$ 3.6 \, \text{atm} \times V_1 = P_{\text{dry gas (new)}} \times 2V_1 $

Solving for $P_{\text{dry gas (new)}}$:

$ P_{\text{dry gas (new)}} = \frac{3.6 \, \text{atm}}{2} = 1.8 \, \text{atm} $

4. The vapor pressure of water remains unchanged at $0.4 \, \text{atm}$ because it depends only on the temperature, not on the volume.

5. The new total pressure of the moist gas is the sum of the new pressure of the dry gas and the unchanged vapor pressure of water:

$ P_{\text{moist gas (new)}} = P_{\text{dry gas (new)}} + P_{\text{water vapor}} = 1.8 \, \text{atm} + 0.4 \, \text{atm} = 2.2 \, \text{atm} $

6. Finally, as requested, we express the new pressure in the form of $\times 10^{-1} \, \text{atm}$ and find the nearest integer:

$ 2.2 \, \text{atm} = 22 \times 10^{-1} \, \text{atm} $

Therefore, the nearest integer is 22.

From ideal gas equation we know

PV = nRT

$ \Rightarrow PV = {W \over M}RT$

$ \Rightarrow P = {W \over V}\,.\,{{RT} \over M}$

$ \Rightarrow P = d\,.\,{{RT} \over M}$ [$\because$ $d = {W \over V}$]

We know, 760 mm of Hg = 1 atm

$\therefore$ 100 mm of Hg = ${{100} \over {760}}$ atm

$\therefore$ Pressure (P) = ${{100} \over {760}}$ atm

Density (d) = 0.46

R = 0.082 L atm K^$-$1 mol^$-$1

T = (257 + 273) K = 530 K

Putting the values in above equation, we get

${{100} \over {760}} = {{0.46 \times 0.082 \times 530} \over M}$

$\Rightarrow$ M = 152

Given, Mass of ideal gas = 100 gm

Let the molar mass of ideal gas = M

$\therefore$ Number of moles of gas (n) = ${{100} \over M}$

Volume of cylinder (V) = 416 L

Temperature (T) = (27 + 273)K = 300 K

Pressure (P) = 1.5 bar

R = 0.083 L bar K^$-$1 mol^$-$1

Using ideal gas equation,

PV = nRT

$ \Rightarrow 1.5 \times 416 = {{100} \over M} \times 0.083 \times 300$

$\Rightarrow$ M = 4

$\mathrm{PV}=\mathrm{nR} T$

$ \mathrm{V}=\frac{2 \times 0.083 \times 300}{2 \times 1}=24.9 \text { litre } $

$\therefore$ Volume of the gas adsorbed per gram of the adsorbent

$ \begin{aligned} &=\frac{24.9}{2.5}=9.96 \mathrm{~L} \\\\ &=9960 ~\mathrm{ml} \end{aligned} $

A nitrogen tank of fixed volume used where number of moles of nitrogen is fixed.

$\therefore$ V = constant

n = constant

R = constant

From ideal gas equation,

PV = nRT

$\Rightarrow$ P $\propto$ T [As V, n, R = constant]

Here, initially P₁ = 30 atm, T₁ = 300 K

Finally, P₂ = ?, T₂ = 318 K

$\therefore$ ${{{P_1}} \over {{P_2}}} = {{{T_1}} \over {{T_2}}}$

$ \Rightarrow {{30} \over {{P_2}}} = {{300} \over {318}}$

$ \Rightarrow {P_2} = 31.8 \simeq 32$

Both gas A and Hydrogen (H₂) gas have same volume at same pressure. Let both 's volume is V and pressure P.

For gas A :

Pressure = P

Temperature (T) = 300 K

Volume = V

Mass = 3 g

Molar mass = M gm/mol

using ideal gas equation,

PV = nRT

$ \Rightarrow PV = {3 \over M} \times R \times 300$ ..... (1)

For Hydrogen,

Pressure = P

Temperature (T) = 200 K

Volume = V

Mass = 0.2 g

Molar mass = 2 gm/mol

Using ideal gas equation,

$PV = {{0.2} \over 2} \times R \times 200$ ...... (2)

From (1) and (2), we get

${3 \over M} \times R \times 300 = {{0.2} \over 2} \times R \times 200$

$\Rightarrow$ M = 45

Weight of empty LPG cylinder = 14.8 kg

Weight of full LPG cylinder = 29 kg

$\therefore$ Weight of gas = 29 $-$ 14.8 = 14.2 kg

If weight of full LPG cylinder = 23 kg

then weight of gas used = 29 $-$ 23 = 6 kg at ambient temperature.

From ideal gas equation, pV = nRT

or $pV = {{Weight\,of\,solute} \over {Molecular\,mass\,of\,solute}} \times RT$

or $pV = {W \over M} \times RT$

Applying ideal gas to LPG cylinder when gas is full,

$pV = nRT$

$3.47\,atm \times V = {{14.2kg} \over M} \times RT$ .... (i)

Applying ideal gas to LPG cylinder when gas is reduced to 23 kg at ambient temperture,

$pV = nRT$

$p \times V = {{8.2kg} \over M} \times RT$ .... (ii)

Divide Eq. (i) by (ii)

${{3.47 \times V} \over {p \times V}} = {{{{14.2kg\,RT} \over M}} \over {{{8.2kg\, \times RT} \over M}}}$

${{3.47} \over p} = {{14.2} \over {8.2}}$

$ \Rightarrow p = {{3.47 \times 41} \over {71}} = 2.003$ atm

Hence, answer is 2.

${{{P_1}} \over {{T_1}}} = {{{P_2}} \over {{T_2}}} \Rightarrow {{300 \times {{10}^3}} \over {300}} = {{1.2 \times {{10}^6}} \over {{T_2}}}$

$ \Rightarrow {T_2} = 1200$ K

$ \Rightarrow $ ${T_2} = 927^\circ $ C

$n(C{H_4}) = {{PV} \over {RT}}$

$ = {{1 \times 4 \times {{10}^3} \times 1000} \over {0.083 \times 300}}$

Weight of CH₄

$ = {{40 \times 16 \times {{10}^5}} \over {0.083 \times 300}}$ gm

$ = 25.7 \times {10^5}$ gm

$n = {{PV} \over {RT}}$

$ = {{1 \times 20 \times {{10}^{ - 3}}} \over {0.083 \times 273}}$

No. of atoms = $ = {{1 \times 20 \times {{10}^{ - 3}}} \over {0.083 \times 273}} \times 2 \times 6.023 \times {10^{23}}$

$ = 1.06 \times {10^{21}}$

V = 10 L, T = 27$^\circ$ C = 300 K

(m)_methane = 6.4 g, (m)_CO2 = 8.8 g

PV = n_totalRT

P $\times$ 10 $\times$ 10^$-$3 = $\left( {{{6.4} \over {16}} + {{8.8} \over {44}}} \right)$ $\times$ 8.314 $\times$ 300

P $\times$ 10^$-$2 = (0.4 + 0.2) $\times$ 8.314 $\times$ 300

P = 149652 Pa

P = 149.652 KPa $ \approx $ 150 kPa

Moles of O₂ = ${{3.12} \over {32}}$ = 0.0975

Using ideal gas equation : PV = nRT

V = ${{0.0975 \times 0.082 \times 300} \over 1}$ = 2.4 L

$ \therefore $ Volume of O₂(g) adsorbed per gram of the

adsorbent = ${{2.4} \over {1.2}}$ = 2

For 1 mole of a real gas, the van der Waals' equation is,

$\left( {p + {a \over {V_m^2}}} \right)({V_m} - b) = RT$

At very high pressure, the equation becomes,

$p({V_m} - b) = RT$

$ \Rightarrow p{V_m} = RT + pb \Rightarrow {{p{V_m}} \over {RT}} = 1 + {{pb} \over {RT}}$

$ \Rightarrow Z = 1 + {{pb} \over {RT}}$ [$\because$ $Z = {{p{V_m}} \over {RT}}$ = compressibility]

$\therefore$ ${\left( {{{\delta Z} \over {\delta p}}} \right)_T} = 0 + {b \over {RT}} + {b \over {RT}} = {{xb} \over {RT}}$ (Given)

$ \Rightarrow x = 1$

${{{P_1}} \over {{T_1}}} = {{{P_2}} \over {{T_2}}}$

${{35} \over {300}} = {{40} \over {{T_2}}}$

${T_2} = {{40 \times 300} \over {35}}$

$ = 342.86$ K

$ = 69.85^\circ $ C $ \simeq 70^\circ $ C

Given, mass of C₂H₂(g) = 4.75 g

Molecular weight = 26 g/mol

Temperature = 50 + 273 = 323 K

Pressure = 740 torr/mm of Hg

Pressure = ${{740} \over {760}}$ atm

R = 0.0821 L atm mol^$-$1 K^$-$1

Hence, no. of mole n = ${{4.75} \over {26}}$ mol

Formula used, pV = nRT (ideal gas)

$ \Rightarrow V = {{nRT} \over p} = {{4.75} \over {26}} \times {{0.0821 \times 323} \over {(740/760)}}$

$ = {{96314.078} \over {19240}} = 5.0059$ L = 5 L

For gas when temperature and number of moles are constant then we can apply Boyle's law.

According to Boyle's law,

P₁V₁ = P₂V₂

Given, P₁ = 48 $ \times $ 10^–3 bar

V₁ = ${{4 \over 3}\pi {{\left( 3 \right)}^3}}$

V₂ = ${{4 \over 3}\pi {{\left( 12 \right)}^3}}$

P₂ = ?

$ \therefore $ P₂ = ${{{P_1}{V_1}} \over {{V_2}}}$

= ${{48 \times {{10}^{ - 3}} \times {{\left( 3 \right)}^3}} \over {{{\left( {12} \right)}^3}}}$

= 7.5 $ \times $ 10^-4 = 750 $ \times $ 10^-6 bar

NaClO₃(s) + Fe(s) $ \to $ NaCl(s) + FeO(s) + O₂(g)

moles of NaClO₃ = moles of O₂

moles of O₂ = ${{PV} \over {RT}}$ = ${{1 \times 492} \over {0.082 \times 300}}$ = 20 mol

mass of NaClO₃ = 20 $ \times $ 106.5 = 2130 g

Aluminium reacts with sulphuric acid to form aluminium sulphate and hydrogen.

$\mathop {2Al}\limits_{\left( {{{5.4} \over {27}} = 0.2\,mol} \right)} + \mathop {3{H_2}S{O_4}}\limits_{\left( {{{50 \times 5} \over {1000}} = 0.25\,mol} \right)} \buildrel {} \over \longrightarrow A{l_2}{(S{O_4})_3} + 3{H_2}$

H₂SO₄ is limiting reagent and moles of H₂(g) produced = 0.25 mol

Using ideal gas equation,

pV = nRT

$ \Rightarrow $ $V = {{0.25 \times 0.082 \times 300} \over {1\,atm}} = 6.15\,L$