Gaseous State

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.

(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.

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.

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 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.

In typical scenarios, especially when discussing gases in everyday sized containers, the effect of gravity on the distribution of pressure within the gas is negligible. This is because gases have low density and high molecular motion which causes the gas molecules to distribute themselves relatively evenly in the available volume. Thus, any influence of gravity is generally not apparent, making the pressure essentially uniform throughout the gas in the container. The main factor influencing the pressure exerted by a gas is described by the ideal gas law, $PV = nRT$, where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin.

This law implies that the pressure is dependent on the temperature, volume, and number of gas molecules but does not include gravity as a factor. Only in very tall containers, or in cases dealing with significantly dense gases under strong gravitational fields (such as on massive planets), does the effect of gravity on the distribution of a gas in a container become significant enough to create a gradient in pressure from top to bottom. For most practical purposes and common situations, gravity's effect on gas pressure in a container can be ignored.

Therefore, the statement is FALSE under typical conditions involving everyday applications and container sizes.