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

The diffusion coefficient (D) is proportional to the mean free path (λ) and the mean speed (U_mean), so we can write

$D ∝ λU_{\text{mean}}$

The mean free path (λ) is given by

$λ = \frac{RT}{\sqrt{2} N_0 σP}$

hence

$λ ∝ \frac{T}{P}$

The mean speed (U_mean) is given by

$U_{\text{mean}} = \sqrt{\frac{8RT}{πM}}$

hence

$U_{\text{mean}} ∝ \sqrt{T}$

Therefore,

$D ∝ \frac{T^{3/2}}{P}$

The change in the diffusion coefficient would then be given by :

$\frac{(DC)_2}{(DC)_1} = \frac{P_1}{P_2} \cdot \left(\frac{T_2}{T_1}\right)^{3/2}$

Substituting $P_2 = 2P_1$ and $T_2 = 4T_1$ into the equation, we get :

$\frac{(D)_2}{(D)_1} = \frac{1}{2} \cdot (4)^{3/2} = 4$

So, the diffusion coefficient of the gas increases 4 times. Hence, x = 4.

Given: External pressure $\left(\mathrm{P}_{\mathrm{ext}}\right)=1 \mathrm{~atm}$

Number of mole of helium $\left(n_{\mathrm{He}}\right)=0.1 \mathrm{~mol}$

No. of mole of unknown compound

$\left(n_{\text {unknown compound }}\right)=1.0 \mathrm{~mol}$

Vapour pressure of unknown compound

$\left(p_{\text {unknown }}^0\right)=0.68 \mathrm{~atm}$

Temperature of the mixture $0^{\circ} \mathrm{C}=273 \mathrm{~K}$

To Find: The volume of gas (in litre) $=v_{\text {gas }}$

Formula: (i) Vapour pressure of helium $\left(\mathrm{P}_{\mathrm{He}}\right)=$

$P_{\text {ext }}-P_{\text {unknown compound }}$

(ii) $\mathrm{V}_{\mathrm{He}}=\frac{n_{\mathrm{He}} \times \mathrm{R} \times \mathrm{T}}{\mathrm{P}_{\mathrm{He}}}$

Since, the evacuated vessel (with fitted) piston in equilibrium with its surroundings. Hence, external pressure (or pressure outside the vessel) is equal to pressure inside the vessel.

$ \begin{aligned} & \mathrm{P}_{\text {ext }}=\mathrm{P}_{\text {internal }}=\mathrm{P}_{\mathrm{T}} \\\\ & \mathrm{P}_{\text {ext }}=\mathrm{P}_{\mathrm{He}}+\mathrm{P}_{\text {unknwon compound }} \\\\ & 1 \mathrm{~atm}=\mathrm{P}_{\mathrm{He}}+0.68 \mathrm{~atm} \\\\ & \mathrm{P}_{\mathrm{He}}=0.32 \mathrm{~atm} \\\\ \end{aligned} $

$\left[\mathrm{P}_{\mathrm{He}} \mathrm{P}_{\text {unknown compound }}\right.$ are partial pressures of helium and unknown gas respectively]

According to ideal gas equation:

$ \begin{aligned} \mathrm{P}_{\mathrm{He}} \times \mathrm{V}_{\mathrm{He}} & =n_{\mathrm{He}} \times \mathrm{R} \times \mathrm{T} \\\\ \mathrm{V}_{\mathrm{He}} & =\frac{0.1 \times 0.0821 \times 273}{0.32} \\\\ \mathrm{~V}_{\mathrm{He}} & =7.004 \mathrm{~L} \end{aligned} $

Let's start by writing down the equations for the root mean square (rms) speed and the most probable speed. The root mean square speed $ v_{rms} $ of a gas with molecular weight $ M $ at a temperature $ T $ in Kelvin is given by the formula:

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

where $ k $ is the Boltzmann constant.

The most probable speed $ v_{mp} $ of a gas is given by:

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

According to the problem statement, at 400 K, the $ v_{rms} $ speed of gas X (with molecular weight $ M_X = 40 $ g/mol) is equal to the $ v_{mp} $ of gas Y at 60 K. We set the equations equal to each other:

$ \sqrt{\frac{3k \times 400}{40}} = \sqrt{\frac{2k \times 60}{M_Y}} $

We simplify this equation. First, we can cancel $ k $ from both sides:

$ \sqrt{\frac{3 \times 400}{40}} = \sqrt{\frac{2 \times 60}{M_Y}} $

Simplify further:

$ \sqrt{\frac{1200}{40}} = \sqrt{\frac{120}{M_Y}} $

$ \sqrt{30} = \sqrt{\frac{120}{M_Y}} $

Squaring both sides gives:

$ 30 = \frac{120}{M_Y} $

Rearrange to solve for $ M_Y $:

$ M_Y = \frac{120}{30} = 4 $

So, the molecular weight of gas Y is 4 g/mol.

The rate of diffusion depends on the following factors.

r $\propto$ P and r $\propto$ $\sqrt {1/M} $

Taking these together, we get

${{{r_2}} \over {{r_1}}} = {{{P_2}} \over {{P_1}}}{\left( {{{{M_1}} \over {{M_2}}}} \right)^{1/2}}$

or ${{{n_1}} \over {{t_1}}} \times {{{t_2}} \over {{n_2}}} = {{{P_1}} \over {{P_2}}} \times \sqrt {{{{M_2}} \over {{M_1}}}} $

or ${1 \over {38}} \times {{57} \over 1} = {{0.8} \over {1.6}} \times \sqrt {{{{M_2}} \over {28}}} $

or ${M_2} = {{57 \times 57} \over {38 \times 38}} \times {{1.6 \times 1.6} \over {0.8 \times 0.8}} \times 28 = 252$

Let the molecular formula of the unknown compound be XeF_n. We will have

${M_{xe}} + n{M_f} = 252$ g mol^$-$1 i.e. $[131 + n(19)]$ g mol^$-$1 = 252 g mol^$-$1

$n = {{252 - 131} \over {19}} = 6.36 \simeq 6$

Hence, the molecular formula of the gas is $Xe{F_6}$.

Weight of butane gas in filled cylinder = 29 $-$ 14.8 kg = 14.2 kg

$\Rightarrow$ During the course of use, weight of cylinder reduces to 23.2 kg

$\Rightarrow$ Weight of butane gas remaining now = 23.2 $-$ 14.8 = 8.4 kg

Also, during use, V (cylinder) and T remains same.

Therefore, ${{{p_1}} \over {{p_2}}} = {{{n_1}} \over {{n_2}}}$

$ \Rightarrow {p_2} = \left( {{{{n_2}} \over {{n_1}}}} \right){p_1} = \left( {{{8.4} \over {14.2}}} \right) \times 2.5$ [Here, ${{{n_2}} \over {{n_1}}} = {{{w_2}} \over {{w_1}}}$]

= 1.48 atm

Also, pressure of gas outside the cylinder is 1.0 atm.

$ \Rightarrow pV = nRT$

$ \Rightarrow V = {{nRT} \over p} = {{(14.2 - 8.4) \times {{10}^3}} \over {58}} \times {{0.082 \times 30} \over 1}L$

= 2460 L = 2.46 m³

Partial pressure of a gas in a mixture = Mole fraction of gas $\times$ Total pressure

$\therefore$ ${P_{He}} = {X_{He}} \times P = {4 \over 5} \times 20 = 16$ bar

${P_{C{H_4}}} = {X_{C{H_4}}} \times P = {1 \over 5} \times 20 = 4$ bar

${r_{He}} = {{k\,.\,{P_{He}}} \over {\sqrt {{M_{He}}} }} = {{k\,.\,16} \over {\sqrt 4 }} = 8k$

${r_{C{H_4}}} = {{k\,.\,{P_{C{H_4}}}} \over {\sqrt {{M_{C{H_4}}}} }} = {{k \times 4} \over {\sqrt {16} }} = k$

$\therefore$ Composition of the mixture (He : CH₄) effusion out = 8k : k = 8 : 1.

According to gas equation,

$PV = nRT$ ; $n = {{2 \times {{10}^{21}}} \over {6.02 \times {{10}^{23}}}}$

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

$ = {{7.57 \times {{10}^3}\,N{m^{ - 2}} \times 1 \times {{10}^{ - 3}}\,{m^3}} \over {{{2 \times {{10}^{21}}} \over {6.02 \times {{10}^{23}}}}\,mol \times 8.314\,J\,(Nm)\,mo{l^{ - 1}}\,{K^{ - 1}}}}$

$ = 274.2$ K

RMS velocity,

$u = \sqrt {{{3RT} \over M}} = \sqrt {{{3 \times 8.314 \times 274.2} \over {28 \times {{10}^{ - 3}}}}} $

$ = 494.2$ ms^$-$1

Most probable velocity = 0.82 $\times$ u

= 494.2 $\times$ 0.82 ms^$-$1 = 405.2 ms^$-$1

${p_{{H_2}}} + {p_g} = 6.0$ atm

where p_g is the pressure exerted by the unknown gas.

${p_{{H_2}}} = {{nRT} \over V} = {{0.7 \times 0.0821 \times 300} \over 3} = 5.747$ atm.

$\therefore$ ${p_g} = 6.0 - 5.747 = 0.253$ atm.

Number of moles of unknown gas

$ = {{{p_g}\,.\,V} \over {RT}} = {{0.253 \times 3} \over {0.0821 \times 300}} = 0.0308$

Rate of effusion of

${H_2} = {{0.7} \over {20}} = 0.035$ mol min^$-$1

Rate of effusion of unknown gas

$ = {{0.0308} \over {20}} = 0.00154$ mol min^$-$1

According to Graham's Law of effusion

${{{M_g}} \over {{M_{{H_2}}}}} = {{{{({r_{{H_2}}})}^2}} \over {{{({r_g})}^2}}} = {{{{(0.035)}^2}} \over {{{(0.00154)}^2}}} = 516.5$

$\therefore$ ${M_g} = 516.5 \times 2 = 1033$ g mol^$-$1

2NO + O₂ $\to$ 2NO₂ $\to$ N₂O₄

Number of moles of NO = ${{PV} \over {RT}}$

$ = {{1.053 \times 250} \over {0.0821 \times 300 \times 1000}} = 0.0107$

Number of moles of

${O_2} = {{0.789 \times 100} \over {0.0821 \times 300 \times 1000}} = 0.0032$

Now 2 moles of NO need 1 mole of O₂ for conversation into NO₂.

$\therefore$ 0.0032 moles of O₂ react with NO

= 0.0064 moles

NO left unreacted

= 0.0107 $-$ 0.0064 = 0.0043 mol

Total volume of the vessels

= 250 + 100 = 350 ml

Oxygen will be completely converted into NO₂ and NO₂ will then be completely converted into N₂O₄ (dimer) which becomes solid at 262 K; hence at 220 K, N₂O₄ is in solid state and only NO is present in gaseous state. Thus the whole volume (250 + 100 = 350 ml) of 350 ml is occupied by NO that has been left unreacted.

Therefore the pressure, P of NO gas = ${{nRT} \over V}$

$ = {{0.0043 \times 0.082 \times 220} \over {0.350}} = 0.221$ atm

Mol. mass of acetylene (C₂H₂) = 26 ; T = 50 + 273 = 323 K

$\therefore$ 5 g acetylene $ = {5 \over {26}}$ moles. Let V be the volume occupied by 5 g of C₂H₂

Applying PV = nRT, we get

${{740} \over {760}} \times V = {5 \over {26}} \times 0.082 \times 323$

or $V = {{5 \times 0.082 \times 323 \times 760} \over {26 \times 740}}L = 5.23\,L$.

To find the total kinetic energy ratio for oxygen and hydrogen at a given temperature, we first note that kinetic energy (average) of a molecule can be represented using the formula for translational kinetic energy of an ideal gas, which is given by:

$ KE_{avg} = \frac{3}{2} k_B T $

where $ k_B $ is the Boltzmann constant and $ T $ is the temperature in Kelvin.

For a given number of molecules $ N $, the total kinetic energy $ KE_{total} $ can be expressed as:

$ KE_{total} = N KE_{avg} $

However, in order to find $ N $, which is the number of molecules, we need the number of moles, since $ N = nN_A $ where $ n $ is the number of moles and $ N_A $ is Avogadro’s number. We calculate $ n $ by dividing the mass of the gas $ m $ by its molar mass $ M $. This gives:

$ n = \frac{m}{M} $

For hydrogen (H₂), the molar mass $ M_{H_2} = 2 $ g/mol and for oxygen (O₂), the molar mass $ M_{O_2} = 32 $ g/mol. Given that both gases have a mass of 8 grams each, we can calculate the moles for each:

$ n_{H_2} = \frac{8 \text{ g}}{2 \text{ g/mol}} = 4 \text{ moles} $

$ n_{O_2} = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ moles} $

The number of molecules $ N $ for each gas becomes:

$ N_{H_2} = n_{H_2} N_A = 4 N_A $

$ N_{O_2} = n_{O_2} N_A = 0.25 N_A $

Even though the mass of each gas is the same, the number of moles (and therefore the number of molecules) is different. However, the total kinetic energy for any ideal gas sample is still dependant on the temperature and the total number of molecules. Therefore, for each gas, substituting from the kinetic energy formula, we get:

$ KE_{total, H_2} = N_{H_2} KE_{avg} = 4 N_A \times \frac{3}{2}k_B T $

$ KE_{total, O_2} = N_{O_2} KE_{avg} = 0.25 N_A \times \frac{3}{2}k_B T $

The ratio of total kinetic energies $(H_2/O_2)$ thus becomes:

$ \frac{KE_{total, H_2}}{KE_{total, O_2}} = \frac{4 N_A \times \frac{3}{2}k_B T}{0.25 N_A \times \frac{3}{2}k_B T} = \frac{4 N_A}{0.25 N_A} = 16 $

Even though both samples are at the same temperature and have the same mass, hydrogen has more molecules contributing to its kinetic energy due to its significantly lower molecular weight compared to oxygen. Hence, under these conditions, the ratio of the total kinetic energy of hydrogen to oxygen is $ 16:1 $.

Let V₁ be the volume of H₂ in the cylinder at NTP. According to gas equation,

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$

$\therefore$ ${{1 \times {V_1}} \over {273}} = {{20 \times 2.82} \over {300}}$

or ${V_1} = {{20 \times 2.82 \times 273} \over {300}}$

= 51.324 L = 51324 mL

Volume of H₂ left in the cylinder = Volume of cylinder = 2820 mL

Actual volume transferred to balloons

= 51324 $-$ 2820 = 48504 mL

Radius of balloon (r) = 21/2 = 10.5 cm

Volume of each balloon = ${4 \over 3}\pi {r^3}$

$ = {4 \over 3} \times 3.142 \times {(10.5)^3}$

$ = 4849.67 = 4850$ mL

Number of balloons which can be filled up

$ = {{48504} \over {4850}} = 10$

The problem posed involves determining the value of PV (pressure-volume product) for a given amount of an ideal gas, specifically 5.6 liters at Normal Temperature and Pressure (NTP). To solve this, we will make use of the ideal gas law, the definition of NTP conditions, and the concept of moles in chemistry.

The ideal gas law is given by:

$ PV = nRT $

where:

• P is the pressure of the gas,
• V is the volume of the gas,
• n is the number of moles of the gas,
• R is the ideal gas constant,
• T is the temperature of the gas.

Normal Temperature and Pressure (NTP) are defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm. According to the conditions set for NTP, the volume of 1 mole of an ideal gas is 22.4 liters.

To find out how many moles are present in 5.6 liters of gas at NTP, we use the relation between the volume of one mole of gas and the provided volume:

$ n = \frac{V}{22.4 \text{ liters/mole}} $

$ n = \frac{5.6 \text{ liters}}{22.4 \text{ liters/mole}} = 0.25 \text{ moles} $

Substituting this back into the ideal gas equation:

$ PV = (0.25 \text{ moles}) \cdot R \cdot T $

At NTP, since the temperature (T) is taken as 273.15 K, and the pressure is 1 atm, the specific value of R can usually be ignored in this context because we are asked for the expression of PV in terms of RT. Therefore:

$ PV = 0.25 RT $

Thus, the value of PV for 5.6 liters of an ideal gas at NTP is 0.25 RT.

The rate of diffusion of a gas is inversely proportional to both the pressure of the gas and the square root of its molecular mass. This relationship is quantitatively expressed in Graham’s Law of Effusion. According to this law, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular mass, mathematically expressed as:

$ R \propto \frac{1}{\sqrt{M}} $

Where R is the rate of diffusion or effusion and M is the molecular mass of the gas.

The rate of diffusion is also influenced by the pressure of the gas. At higher pressures, the rate of diffusion decreases. However, the relationship between gas pressure and the rate of diffusion is not as straightforward as the relationship with molecular mass, and it isn't as commonly discussed in simple terms as is Graham’s Law. Typically, the effect of pressure on the rate of diffusion is more significant in specific contexts, such as under varying atmospheric conditions.

In fundamental terms, a higher pressure implies more gas molecules are present, leading to increased collision frequency among them, which can slow down the overall rate at which individual molecules spread out or move through a medium.

${U_{rms}} = \sqrt {{{3RT} \over M}} = \sqrt {{{3 \times 8.314 \times {{10}^7} \times 293} \over {48}}} $

$ = 3.9 \times {10^4}$ cm sec^$-$1

To find the total energy of one mole of an ideal monatomic gas, we need to use the formula related to the internal energy of the gas. For an ideal monatomic gas, the internal energy ($U$) is given by the expression:

$ U = \frac{3}{2} nRT $

where:

• $n$ is the number of moles of the gas,
• $R$ is the ideal gas constant, and
• $T$ is the temperature in Kelvin.

Given that $n = 1$ mole and the temperature is $27^\circ C$, we first need to convert this temperature to Kelvin. The conversion from Celsius to Kelvin is done by adding 273:

$ T = 27 + 273 = 300 \, \text{K} $

The ideal gas constant $R$ can be used in various units, but since we want the energy in calories, we will use the value of $R = 2 \, \text{cal/mol}\cdot\text{K}$. Plugging these values into the formula, we get:

$ U = \frac{3}{2} \times 1 \times 2 \times 300 \, \text{cal} $

$ U = \frac{3}{2} \times 600 \, \text{cal} $

$ U = 900 \, \text{cal} $

Thus, rounding to an appropriate number of significant figures, the total energy of one mole of an ideal monatomic gas at 27°C is approximately 900 calories.

The difference between the molar heat capacities at constant pressure ($C_p$) and constant volume ($C_v$) for an ideal gas is a constant value given by the gas constant $R$. This relationship is expressed mathematically as:

$ C_p - C_v = R $

Here, $R$ is the universal gas constant, with a value of approximately 8.314 J/(mol·K).

The reason behind this relationship lies in the energy needed to perform work against the external pressure when the gas is heated at constant pressure, which does not happen in the case of heating at constant volume. When an ideal gas is heated at constant pressure, not only does the internal energy of the gas increase (which is also the case at constant volume) but some of the energy is also used to do work on the surroundings by expanding, which contributes to the larger value of $C_p$ compared to $C_v$. This difference is equal to the amount of work done per unit increase in temperature, which is quantitatively equivalent to the gas constant $R$.

Therefore, the expression indicates that the difference in heat capacity (which essentially is the difference in the ability to store thermal energy under different conditions) is fundamentally linked to the work done by the gas during expansion at constant pressure. The equation $C_p - C_v = R$ holds true for an ideal gas because it assumes no interactions between the molecules and that all processes are reversible.