Gaseous State

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.

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.