Gaseous State

Partial Pressure Ratio of Nitrogen to Oxygen :

The partial pressure of a gas is calculated using its mole fraction. The mole fraction is the ratio of the number of moles of the gas to the total number of moles of all gases.

For nitrogen ($ \text{N}_2 $), the mole fraction $\frac{n_{\text{N}_2}}{n_{\text{N}_2} + n_{\text{O}_2} + n_{\text{Ar}}}$ can be simplified as $\frac{70/28}{70/28 + 27/32 + 3/40}$.

For oxygen ($ \text{O}_2 $), the mole fraction is $\frac{27/32}{70/28 + 27/32 + 3/40}$.

Therefore, the ratio of the partial pressure of nitrogen to oxygen is :

$ \frac{P_{\text{N}_2}}{P_{\text{O}_2}} = \frac{\text{mole fraction of } \text{N}_2}{\text{mole fraction of } \text{O}_2} = \frac{70/28}{27/32} = 2.96 $

Partial Pressure Ratio of Oxygen to Argon :

For argon ($ \text{Ar} $), the mole fraction is $\frac{3/40}{70/28 + 27/32 + 3/40}$.

The ratio of the partial pressure of oxygen to argon is :

$ \frac{P_{\text{O}_2}}{P_{\text{Ar}}} = \frac{\text{mole fraction of } \text{O}_2}{\text{mole fraction of } \text{Ar}} = \frac{27/32}{3/40} = 11.25 $

The van der Waals constant 'a' is a measure of the strength of the intermolecular forces in a gas. Larger molecules and molecules with stronger intermolecular forces will have larger 'a' constants.

Here, we are comparing argon (Ar), methane (CH₄), water (H₂O), and benzene (C₆H₆).

Argon (Ar) is a noble gas and is monoatomic. Methane (CH₄) is a small molecule with weak intermolecular forces. Water (H₂O) is a small molecule but has strong hydrogen bonding, making its intermolecular forces stronger than those in methane. Benzene (C₆H₆) is a larger molecule than the others and has relatively strong intermolecular forces due to its larger size and polarizability.

So, in the increasing order of van der Waals constant 'a', it should be:

Argon (Ar) < Methane (CH₄) < Water (H₂O) < Benzene (C₆H₆)

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}$ [$\because$ $d = {W \over V}$]

For a fixed amount of gas at a fixed temperature, M and T is constant.

$\therefore$ $P \propto d$

So graph between pressure (P) and density (d) is a straight line.

Also, ${P \over d} = {{RT} \over M}$

$ \Rightarrow {P \over d} \propto T$ [as R, M = constant]

$\therefore$ When temperature T increases then slope of graph between P and d increases.

As T₃ > T₂ > T₁ then slope of T₃ will be highest then T₂ and then T₁.

So, graph (B) is the right graph.

Weight of empty glass vessel = 40 gm

Weight of glass vessel filled with liquid = 135 gm

$\therefore$ Weight of liquid = 135 $-$ 40 = 95 gm

Given density of liquid = 0.95 gm ml^$-$1

$\therefore$ Volume of liquid $={{95} \over {0.95}} = 100$ ml

Weight of glass filled with ideal gas = 40.5 gm

$\therefore$ Weight of gas = 40.5 $-$ 40 = 0.5 g

Let the Molar mass = M

$\therefore$ Moles of gas $ = {{0.5} \over M}$

$\therefore$ Now applying ideal gas equation,

pV = nRT

$ \Rightarrow 0.82 \times {{100} \over {1000}} = {{0.5} \over M} \times 0.082 \times 250$

$\Rightarrow$ M = 0.5 $\times$ 250 = 125 g/mol

Root mean square (rms) velocity of N₂ molecule (u) = $\sqrt {{{3RT} \over M}} $.

For N₂ molecule, molecular mass, M = 28. Therefore,

$u = \sqrt {{{3RT} \over {28}}} $ ..... (1)

When the temperature becomes doubled, that is, 2T. Let the new root mean square (rms) velocity of N-atom be u'. For N-atom, the molecular mass, M = 14. Therefore,

$u' = \sqrt {{{3RT} \over {14}}} $....... (2)

On dividing Eq. (1) by Eq. (2), we get

${u \over {u'}} = \sqrt {{1 \over 4}} $

$ = {1 \over 2} \Rightarrow u' = 2u$

According to van der Waals' equation, for one mole of a gas

$\left( {p + {a \over {{V^2}}}} \right)(V - b) = RT$ ...... (1)

At high pressure, ${a \over {{V^2}}}$ can be neglected

So, $p + {a \over {{V^2}}} \approx p$ ...... (2)

From Eqs. (1) and (2), we get

$p(V - b) = RT$

$pV - pb = RT$

$pV = RT + pb$

Dividing both the sides by RT, we get

${{pV} \over {RT}} = 1 + {{pb} \over {RT}}$

$Z = 1 + {{pb} \over {RT}}$

To solve this problem, we'll start by using the ideal gas law and the formula for the root mean square (rms) velocity of a gas.

The ideal gas law is given by:

$ PV = nRT $

where $ P $ is the pressure, $ V $ is the volume, $ n $ is the number of moles, $ R $ is the universal gas constant, and $ T $ is the temperature. The rms velocity ($ v_{rms} $) of a gas is given by:

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

where $ M $ is the molar mass of the gas. We need the rms velocities of gases $ X $ and $ Y $.

First, let's find the number of moles of each gas. For gas $ X $, let’s denote the molar mass as $ M_X $. The number of moles is:

$ n_X = \frac{10 \, \text{g}}{M_X} $

For the initial gas $ X $ at $ 300 \, \text{K} $ and $ 2 \, \text{atm} $, we can write:

$ P_X V = n_X RT $

When gas $ Y $ is added, the total pressure becomes $ 6 \, \text{atm} $. Let the molar mass of $ Y $ be $ M_Y $. The number of moles of gas $ Y $ added is:

$ n_Y = \frac{80 \, \text{g}}{M_Y} $

Now, the total number of moles in the vessel is $ n_X + n_Y $, and this total exerts a pressure of $ 6 \, \text{atm} $ at the same volume and temperature:

$ P_{total} V = (n_X + n_Y) RT $

Given the total pressure is $ 6 \, \text{atm} $, solving for the volume$ V $ and equating, we get two equations:

$ 2V = \frac{10}{M_X} RT \quad ......\text{(i)} $

$ 6V = \left( \frac{10}{M_X} + \frac{80}{M_Y} \right) RT \quad ......\text{(ii)} $

Dividing (ii) by (i) and solving for $ \frac{80}{M_Y} $, we get:

$ \frac{80}{M_Y} = 2 \frac{10}{M_X} $

$ \frac{80}{M_Y} = \frac{20}{M_X} $

$ M_Y = 4 M_X $

Now, applying the formula for the rms velocity, the ratio of rms velocities of $ X $ and $ Y $ is:

$ \frac{v_{rms, X}}{v_{rms, Y}} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{\frac{4 M_X}{M_X}} = 2 $

So, the ratio of root mean square velocities of $ X $ and $ Y $ is $ 2:1 $.

Thus, the correct answer is Option D: $ 2:1 $.

The ratio of most probable, the average and the root mean square speeds for this graph is 1 : 1.128 : 1.224

But the graph in question is completely symmetrical. Therefore, the most probable and the average speed will be same here but root mean square speed will be greater than average speed. So, the correct ratio is 1 : 1 : 1.224.

The van der Waals equation is

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

but it is given that b = 0. So, the equation reduces to

$\left( {p + {a \over {{V^2}}}} \right)V = RT \Rightarrow pV = - {a \over V} + RT$

Comparing it with a straight line equation we get slope as $-$a. Calculating the slope, we get

${{24.6 - 20.1} \over {3.0 - 0}} = 1.5 \Rightarrow a = 1.5$

The van der Waals equation is a modified version of the ideal gas law which accounts for the non-ideal behavior of real gases. This adjustment is done through two correction terms, each addressing a specific factor where real gases deviate from ideal behavior:

• Intermolecular attraction
• Volume occupied by the gas particles themselves

The van der Waals equation is given by:

$\left(P + \frac{{a n^2}}{{V^2}}\right) \left(V - nb\right) = nRT$

Let's break down the components:

1. $\left(P + \frac{{a n^2}}{{V^2}}\right)$: This term adds a correction to the pressure (P) of the gas. The parameter $a$ provides a correction for the intermolecular forces. As these forces are attractive, they effectively reduce the pressure exerted by the gas. Thus, $\frac{{a n^2}}{{V^2}}$ is added to the actual pressure to account for this decrease due to attraction.
2. $\left(V - nb\right)$: The corrected volume term where $b$ accounts for the volume occupied by the gas particles themselves, which is otherwise not considered in the ideal gas law. The subtraction by $nb$ (where $n$ is the number of moles of the gas and $b$ is a constant related to the volume occupied by one mole of gas particles) corrects for the volume unavailable to the gas particles for movement.

From the options provided:

• Option A ($nb$) and Option D ($-nb$) relate to the volume correction for the space that the gas molecules occupy.
• Option B ($\frac{{a n^2}}{{V^2}}$) and Option C ($- \frac{{a n^2}}{{V^2}}$) relate to the intermolecular forces.

The enhancement factor $\frac{{a n^2}}{{V^2}}$ is added to the pressure to counteract the lowering effect of the attractions on the measured pressure, thus the term correcting for the attractive forces and increasing the effective pressure would be positively stated, not negatively. Therefore:

Option B ($\frac{{a n^2}}{{V^2}}$) is the correct answer, as it accounts for the attractive forces in the modification of the ideal gas law in the van der Waals equation.

(A) $Z = P{V_m}/RT$ at high pressure and low temperature. The value of Z tends towards infinity at very high pressure such as 200 bar.

Equation $(P + a{n^2}/{V^2})(V - nb) = nRT$ reduces to $P(V - nb) = nRT$.

(B) For hydrogen gas value of Z = 1 at P = 0 and it increase continuously on increasing pressure. At such low pressure the gas behaves ideally, the van der Waals equation is reduced to :

$PV = nRT$

(C) As can be seen from the plot of compressibility factor (Z) versus pressure (in bar), the compressibility factor for carbon dioxide gas is less than 0 at 1 atm (or 1.013 bar). This is because CO$_2$ molecules have larger attractive forces, under normal conditions.

(D) $Z = P{V_m}/RT$, at very large molar volume Z $\ne$ 1.

The van Der Waals equation for real gas:

$\left( {P + {{a{n^2}} \over {{V^2}}}} \right)(v - nb) = nRT$

All very large molar volume, ${{a{n^2}} \over {{V^2}}}$ almost neglibible, hence, equation is reduced to

$P(v - nb) = nRT$.

The average velocity of a molecule in an ideal gas is related to the temperature, which can be determined using the formula for the root mean square speed (v_rms), derived from the Maxwell-Boltzmann distribution. The formula for v_rms is:

$ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} $

where:

• k is the Boltzmann constant,
• T is the absolute temperature in Kelvin,
• m is the mass of a molecule of the gas.

To find the average velocity at two different temperatures, it is important to note that while the average velocity and the root mean square speed are not exactly the same, they are proportional to the square root of the temperature. Therefore, for the purpose of comparing speeds at two temperatures, we can use the formula:

$ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} $

Where:

• v₁ and v₂ are the speeds at temperatures T₁ and T₂ respectively.

Here, temperatures must be converted to Kelvin:

• T₁ = 27^o C = 27 + 273.15 = 300.15 K
• T₂ = 927^o C = 927 + 273.15 = 1200.15 K

Plugging these values into the proportion formula gives:

$ \frac{v_2}{0.3} = \sqrt{\frac{1200.15}{300.15}} $

Calculating the square root:

$ \sqrt{\frac{1200.15}{300.15}} \approx \sqrt{4} = 2 $

Therefore:

$ v_2 = 0.3 \times 2 = 0.6 \, \text{m/sec} $

Thus, the average velocity of the gas molecules at 927ºC is 0.6 m/sec.

Therefore, the correct answer is: Option A (0.6 m/sec).

Diffusion of a gas is described by Graham's Law, which states that the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its molecular mass. This law is mathematically represented as:

$ \text{Rate of diffusion} \propto \frac{1}{\sqrt{\text{Molecular Mass}}} $

Given this information, let's analyze the options:

• Option A: Directly proportional to its density - This is incorrect because density and molecular mass are related, but the rate of diffusion is inversely related to the square root of molecular mass, not directly proportional to density.
• Option B: Directly proportional to the square root of its molecular weight - This is incorrect. The relationship is inverse; the rate decreases as the molecular weight increases.
• Option C: Inversely proportional to the square root of its molecular weight - This is the correct option. As molecular mass increases, the rate of diffusion decreases in proportion to the inverse of the square root of the molecular mass.
• Option D: Directly proportional to its molecular weight - This is incorrect as it contradicts the principle of Graham's Law.

Thus, the correct answer is Option C: inversely proportional to the square root of its molecular weight.

To find the fraction of the total pressure exerted by hydrogen, we need to use Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is the sum of the partial pressures of each component gas in the mixture. The partial pressure exerted by each gas is directly proportional to its mole fraction in the mixture.

First, let's determine the number of moles of methane (CH₄) and hydrogen (H₂) given that equal weights of each are mixed.

The molar mass of CH₄ is approximately 16 g/mol, and the molar mass of H₂ is approximately 2 g/mol. Let the common weight of each gas be $ w $ grams. Then, the number of moles of methane is given by:

$ n_{CH_4} = \frac{w}{16} \text{ moles} $

The number of moles of hydrogen is:

$ n_{H_2} = \frac{w}{2} \text{ moles} $

Next, we calculate the total number of moles $ n_{total} $ in the mixture:

$ n_{total} = n_{CH_4} + n_{H_2} = \frac{w}{16} + \frac{w}{2} $

$ n_{total} = \frac{w}{16} + \frac{8w}{16} = \frac{9w}{16} \text{ moles} $

To find the fraction of the total pressure exerted by hydrogen, we calculate the mole fraction of hydrogen, which is the ratio of the number of moles of hydrogen to the total number of moles:

$ \text{Mole fraction of } H_2 = \frac{n_{H_2}}{n_{total}} = \frac{\frac{w}{2}}{\frac{9w}{16}} = \frac{16}{18} = \frac{8}{9} $

Dalton's Law implies that the fraction of total pressure exerted by hydrogen is equal to the mole fraction of hydrogen. Therefore, the pressure fraction exerted by hydrogen is:

$ \frac{8}{9} $

Thus, the correct answer is Option B: 8/9.

The average kinetic energy of a gas molecule or atom at a given temperature is described by the kinetic theory of gases. The formula used to calculate the average kinetic energy (KE) of each particle in a gas is given by the equation:

$ KE = \frac{3}{2} kT $

where:

• $ k $ is the Boltzmann constant, and
• $ T $ is the absolute temperature in Kelvin.

This equation reveals that the average kinetic energy of any ideal gas particle depends only on the temperature, and not on the mass of the particles. Hence, at any given temperature, all ideal gas particles have the same average kinetic energy.

Given that both helium (He) and hydrogen molecule (H₂) are treated as ideal gases, at the same temperature of 298 K, the average kinetic energy of a helium atom would therefore be:

$ KE_{\text{He}} = \frac{3}{2} k (298) $

Similarly, for the hydrogen molecule:

$ KE_{\text{H}_2} = \frac{3}{2} k (298) $

It is evident that both expressions are identical, indicating that at 298 K, the average kinetic energy of a helium atom is the same as that of a hydrogen molecule. Despite the difference in mass between helium and hydrogen, conditions of temperature govern the kinetic energy within the framework of kinetic theory for ideal gases.

The correct answer is:

Option C: same as that of a hydrogen molecule

The behavior of real gases is often analyzed to see how closely they follow the ideal gas laws, particularly as these laws are approached under varying conditions of temperature and pressure. The correct answer in context to the temperature at which a real gas behaves most like an ideal gas over a wide range of pressures is not the critical temperature, inversion temperature, or reduced temperature. Instead, it is the Boyle temperature.

To elaborate, let's look at each option:

• Critical temperature is the temperature above which a gas cannot be liquefied by pressure alone. At this temperature, the properties of the gas change significantly, which does not necessarily imply ideal gas behavior.
• Boyle temperature is the specific temperature at which a real gas adheres to Boyle's Law (PV = constant) over a wide range of pressures. At this temperature, the attractive and repulsive forces between molecules balance out in such a way that the gas's P-V behavior (pressure-volume) mimics that of an ideal gas.
• Inversion temperature is related to the Joule-Thomson effect, which describes how the temperature of a real gas changes when it expands or compresses without the exchange of heat with its surroundings. This is different from behaving like an ideal gas over a range of pressures.
• Reduced temperature is a term used in corresponding states theory to correlate the properties of substance under various conditions. It is dimensionless and given by $ T_r = \frac{T}{T_c} $ where $ T $ is the absolute temperature and $ T_c $ is the critical temperature.

Since the question specifically asks about the temperature at which the behavior of a real gas is similar to that of an ideal gas across a broad range of pressures, the answer is Option B: Boyle temperature. This is the temperature where deviations from ideal gas law due to intermolecular forces are minimized in the behavior of real gases, thus aligning more closely to the predictions by the ideal gas law (PV=nRT).

The question deals with the calculation of the ratio of the root mean square velocity to the average velocity of gas molecules. To solve it, we need equations for both these velocities.

The root mean square velocity, $v_{rms}$, is given by the formula:

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

where:

• $R$ is the universal gas constant,


• $T$ is the temperature,


• $M$ is the molar mass of the gas.

The average velocity, $v_{avg}$, can be calculated using the formula:

$ v_{avg} = \sqrt{\frac{8RT}{\pi M}} $

To find the ratio of the root mean square velocity to the average velocity, we take the ratio of these two expressions:

$ \frac{v_{rms}}{v_{avg}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}} $

Simplifying this, we can cancel out $RT/M$ from the numerator and denominator since they are present in both terms:

$ \frac{v_{rms}}{v_{avg}} = \frac{\sqrt{3}}{\sqrt{\frac{8}{\pi}}} = \sqrt{\frac{3\pi}{8}} $

To solve $ \sqrt{\frac{3\pi}{8}} $ numerically, we know that $\pi \approx 3.14159$. Plugging in the values, we get:

$ \sqrt{\frac{3 \times 3.14159}{8}} \approx \sqrt{1.178} \approx 1.085 $

So, the ratio of $v_{rms}$ to $v_{avg}$ is approximately 1.085, which is nearest to Option A, 1.086:1.

Therefore, the correct answer is Option A, 1.086:1.

To find the fraction of the total pressure exerted by oxygen, we need to consider the mole fractions of methane (CH₄) and oxygen (O₂) in the mixture. Here are the steps to solve this problem:

1. Let us assume equal weights, say 'W' grams, of methane and oxygen are mixed.

2. The molecular weights of methane and oxygen are 16 g/mol and 32 g/mol, respectively. The moles of methane (n_CH4) and moles of oxygen (n_O2) in the mixture are then calculated as follows:

• $ n_{CH_4} = \frac{\text{Mass of CH}_4}{\text{Molar mass of CH}_4} = \frac{W}{16} \text{ moles} $
• $ n_{O_2} = \frac{\text{Mass of O}_2}{\text{Molar mass of O}_2} = \frac{W}{32} \text{ moles} $

3. For W grams of each gas, ratio of moles becomes:

• $ \text{The mole ratio of CH}_4 \text{ to O}_2 = \frac{W/16}{W/32} = \frac{32}{16} = 2 \text{ to } 1 $

4. With 2 moles of CH₄ for every 1 mole of O₂, the total moles in the mixture are 3 moles (2 moles CH₄ + 1 mole O₂).

5. According to Dalton's law of partial pressures, the total pressure P in the container is the sum of the partial pressures of the components. Each component's partial pressure is proportional to its mole fraction:

• Mole fraction of O₂, $ X_{O_2} = \frac{n_{O_2}}{n_{total}} = \frac{1}{3} $

6. Hence, the fraction of the total pressure exerted by oxygen would be:

$ \text{Fraction exerted by O}_2 = X_{O_2} = \frac{1}{3} $

Therefore, the correct answer is: Option A - 1/3.