Gaseous State

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

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

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

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

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

(A) When gas molecules collide with each other, high energy molecule transfers some of its energy to less energetic molecule, but the total energy remains conserved. This type of collision between gas molecules is called elastic collision. According to kinetic theory of gas, no loss of energy is observed on collisions and kinetic energy is proportional to absolute temperature of gas.

Option (A) is correct.

(B) Momentum of any particle for example, gas molecules depends upon the mass as well as the velocity of the gas. But when there is huge increase in the mass of the gas, its velocity decreases considerably. Hence, heavier gas molecules collides the walls of container with less impact. The transfer of momentum to the walls is very less.

Option (B) is incorrect.

(C) Maximum number (no.) of gas molecules have speed distributed around the most probable speed ( $v_{\mathrm{mps}}$ ). This speed corresponds to the maximum peak of Maxwell-Boltzmann distribution curve.

There are few molecules which have very high speed (the shaded area under the curve).

Option $(\mathrm{C})$ is correct.

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.

Van der Waals equation is $\left( {P + {{a{n^2}} \over {{V^2}}}} \right)(V - nb) = nRT$, where a is a constant whose value depends on the attraction between the gas molecules and b is the volume that is occupied by one mole of gas.

In option (A), it behaves similar to an ideal gas in the limit of large molar volumes. When the volume is very large then the attraction between the gas molecules can be neglected, so a $\approx$ 0 and the volume occupied by one mole of gas compared to large volume of container will also be neglected. So, b $\approx$ 0. Now, we can neglect a and b and the gas behaves ideally. Thus, option A is correct.

In option (B), it behaves similar to an ideal gas in limit of low pressure and high temperatures. Now, the pressure is very low. So, the attraction between the particles is negligible.

Van der Waals coefficients that are dependent on the identity of the gas but are independent of the temperature. Van der Waals coefficient depends on the attraction between the gas molecules (each of which is characteristic of a gas) and b is the volume that is occupied by one mole of the molecules. So, this option is correct. Since, molecules of Van der Waals gas exert force of attraction on each other; hence, the force which they exert on the walls of the container is less than the force exerted by it if the gas would have been ideal.

Thus, option (D) is correct.

(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

$P(v - nb) = nRT$.

From the given graph, it is clear that, the value of $Z$ decreases as pressure increases.

The expression for Z is given below in equation (i),

$ \mathrm{Z}=\frac{\mathrm{PV}}{n \mathrm{RT}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

The given equation is for van der waals equation, the modification of an ideal gas equation.

$ \left(\mathrm{P}+\frac{a n^2}{\mathrm{~V}^2}\right)(\mathrm{V}-b)=n \mathrm{RT}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

Here, P is the pressure, V denotes volume, R is the universal gas constant, $n$ is moles of gas and T is absolute temperature. The constants a and $b$ represents magnitude of intermolecular attraction and excluded volume respectively.

At higher pressure, the value of P will be large and that of V will be smaller. But, the value of $b$ cannot be ignored as compared to that of V . As V is very small, the term $\left(\frac{a n^2}{\mathrm{~V}^2}\right)$ will have negligible value and it can be ignored with respect to value of $P$ in van der waals equation. Hence, van der waals equation that is equation (ii) can be written as:

$ \mathrm{P}(\mathrm{~V}-b)=n \mathrm{RT} $

Assume $n=1$ for reference. The above expression can be written as:

$ \mathrm{P}(\mathrm{~V}-b)=\mathrm{RT} $

On solving above expression,

$ \mathrm{PV}-\mathrm{P} b=\mathrm{RT} $

On rearranging the above equation,

$ \mathrm{PV}=\mathrm{RT}+\mathrm{Pb} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

$ \begin{aligned} &\text { Divide equation (iii) by RT, }\\ &\frac{\mathrm{PV}}{\mathrm{RT}}=1+\frac{\mathrm{Pb}}{\mathrm{RT}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Assume $n=1$ for reference and $Z$ can be given as:

$ Z=1+\frac{P b}{R T} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

The equation (v) depicts that, the value of $Z$ will increases with pressure linearly if $Z>1$.

According to graph, gas A is showing positive deviation and gas $B$ shows negative deviation.

Therefore, statement A is correct. In case of gas B, the negative deviation shows that the value of $Z$ decreases with increase in $P$. Hence, statement B is incorrect.

For statement C , the graph shows that the gas $C$ behaves like a real gas and the value of $a$ or $b$ will not be equal to zero. Hence, statement $C$ is correct.

For statement D, consider equation (v). The slope can be calculated using equation,

$ y=m x+c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)$

On comparing equation (v) and (vi), $y=\mathrm{Z}$, $x=\mathrm{P}$ and $c=1$.

Therefore, $\quad m=\frac{b}{\mathrm{RT}}$

As $b, \mathrm{R}$ and T are positive, $\frac{b}{\mathrm{RT}}>0$.

Hence, slope will be positive and greater than 0 . Therefore, option D is correct.

The assertion that a mixture of ideal gases forms an "ideal solution" when cooled down to liquid helium temperature (4.22K) requires clarification on several aspects. First, we need to understand the behavior of gases at extremely low temperatures, particularly as they approach liquefaction, and how this correlates to the concept of an "ideal solution."

An ideal gas, by definition, is a hypothetical gas composed of many randomly moving point particles that interact only through elastic collisions and do not exert any other forces on each other. At significantly low temperatures, particularly around 4.22 K (the boiling point of helium), many gases would condense into liquids or solids, depending on their respective boiling and freezing points.

The concept of an "ideal solution" generally applies to the solution phase and is defined as a solution that obeys Raoult's Law at all concentrations for both the solvent and the solute. Raoult's Law states that the partial vapor pressure of each component in a solution is proportional to its mole fraction in the solution. This law is strictly applicable under the assumption of no volume change on mixing and similar intermolecular forces among the components, which is often not precisely true when different substances are involved.

When ideal gases are cooled to low temperatures, such as 4.22 K, and liquefy, the behavior of the resultant liquid may not necessarily align with the principles of an ideal solution. The intermolecular forces in a liquid, particularly at such low temperatures, are significantly stronger and different in nature compared to those in a gaseous state. Moreover, the assumption of similar intermolecular forces and no volume change upon mixing is quite often violated when different gases are involved.

Therefore, it is generally FALSE to claim that cooling a mixture of ideal gases to a temperature as low as 4.22 K to form a liquid would result in an ideal solution. While the gases might mix uniformly, the resulting liquid mixture can exhibit deviations from ideal solution behavior due to changes in intermolecular forces and interactions that are not accounted for by ideal gas or ideal solution theories.

The statement in the question is FALSE, which corresponds to Option B. In the van der Waal's equation, $ (P + \frac{n^2 a}{V^2})(V - nb) = nRT $, the constants 'a' and 'b' have specific roles, but they do not represent the actual volume of the gas molecules directly.

The constant 'a' in the van der Waals equation accounts for the intermolecular forces of attraction between the gas molecules. A higher value of 'a' indicates stronger attraction between the molecules, which affects how closely they can be packed together. The term $ \frac{n^2 a}{V^2} $ adds to the pressure 'P' to compensate for the lowering of pressure due to these attractive forces.

On the other hand, the constant 'b' represents the volume occupied by the gas molecules themselves, known as the excluded volume. It corrects the volume term 'V' in the ideal gas equation $ PV = nRT $, modifying it to $ (V - nb) $ to account for the volume that cannot be occupied by other molecules because it is already taken up by the gas molecules themselves.

Therefore, it is 'b', not 'a', that reflects the actual volume of the gas molecules in the van der Waal's equation.

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.

The statement that the kinetic energy of a molecule is zero at 0°C is FALSE (Option B).

Kinetic energy is directly related to the temperature of an object, but it does not reach zero unless the temperature is at absolute zero, which is -273.15°C or 0 Kelvin. At 0°C (273.15 Kelvin), molecules in any substance still possess kinetic energy. This kinetic energy is due to the motion of the molecules, which continue to vibrate, rotate, or translate depending on their state (solid, liquid, or gas).

The kinetic energy (KE) of an individual molecule can be quantified using the equation:

$ KE = \frac{1}{2} m v^2 $

Where:

• m = mass of the molecule
• v = velocity of the molecule

At any temperature above absolute zero, the velocity v is not zero, so the kinetic energy is also not zero. As long as there is thermal energy (any temperature above absolute zero), molecules will have kinetic energy due to their motion.

Therefore, at 0°C, molecules still have kinetic energy derived from their thermal energy, proving the statement to be false.