Gaseous State

Step 1: Understanding ideal gas behavior.
For gases to behave ideally, the temperature should be high, and the pressure should be low. The given temperature and pressure in option (C) are likely to lead to ideal gas behavior.

Step 2: Analyzing the options.
(A) 100 °C and 10.0 atm: Incorrect — At higher pressures, gases deviate from ideal behavior.
(B) 0 °C and 0.50 atm: Incorrect — Although the pressure is low, the temperature is very low.
(C) 200 °C and 0.50 atm: Correct — This combination is closest to the conditions for ideal gas behavior.
(D) -100 °C and 10.0 atm: Incorrect — Both the low temperature and high pressure result in deviations from ideal gas behavior.

Step 3: Conclusion.
The correct answer is (C) because this temperature and pressure are closest to ideal gas behavior.

Step 1: Using the ideal gas equation.
The ideal gas equation is: where is pressure, is volume, is the number of moles, is the ideal gas constant, and is temperature. To find the density, we rearrange the equation to express as the molar concentration: The molar mass of is 32 g/mol, so the density ( ) is given by: Substituting the given values:

Step 2: Conclusion.
The density of at 300 K and 1.0 atm is 1.299 g L .

To find the molar mass of the gas, we use the relation for the most probable speed in a Maxwell distribution:

where is the Boltzmann constant, is the temperature, and is the mass of one molecule.

Given (from the graph) and .

For molar mass , using , we have:

Solving for :

Substitute values:

Rounded to one decimal place, .

Behavior at Boyle Temperature ( )

The Boyle temperature ( ) is the temperature at which a real gas exhibits ideal gas behavior over an appreciable range of pressure, especially in the low-to-moderate pressure region.

Ideal Gas Behavior ( ): For an ideal gas, the compressibility factor is equal to 1 at all temperatures and pressures.

Definition of : At , the effects of intermolecular attractive forces (which tend to make ) and repulsive forces (which tend to make ) effectively balance out. This is formally defined as the temperature where the second virial coefficient ( ) is zero.

Low Pressure Limit: Because the attractive and repulsive forces balance, the gas behaves ideally (i.e., ) over a wider pressure range compared to any other temperature. Therefore, the vs. curve starts at at and remains close to for low pressure .

High Pressure Limit: As pressure increases significantly, the repulsive forces (due to the finite volume of the gas molecules, represented by the Van der Waals constant ) eventually dominate. This causes the molar volume of the real gas to be greater than that of an ideal gas, leading to . Consequently, the curve must bend upwards.

Graph (B) starts at and remains horizontal or nearly horizontal for an initial pressure range before gradually increasing to at high pressures, which correctly depicts the behavior of a real gas at its Boyle temperature.

The problem involves identifying the one-dimensional Maxwell-Boltzmann distribution of velocities, specifically for the component of velocity , at a given temperature . We are given that is the normalization constant and is the Boltzmann constant. Let's break down the reasoning to find the correct expression among the given options.

Firstly, the Maxwell-Boltzmann distribution for one-dimensional velocities is generally given by:

• ,

where:

• is the mass of the particle,
• is the Boltzmann constant,
• is the absolute temperature,
• is the velocity component in one dimension,
• denotes the exponential function.

Now, let's examine the options:

The first option matches the one-dimensional Maxwell-Boltzmann velocity distribution form. The exponent in the distribution contains a factor of in the denominator, which is typical in the Gaussian like distribution for velocity, representing .

Options 2, 3, and 4 introduce variations that do not match the classical form of the one-dimensional velocity distribution.

Thus, the correct option is indeed:

This justifies that among the given options, the correct expression for the Maxwell-Boltzmann distribution of one-dimensional velocities at temperature is:

• (A) This statement is true for an ideal gas. The internal energy of an ideal gas depends only on temperature, and for an ideal gas, since the energy is a function of temperature alone and not of volume.
• (B) This is incorrect because for an ideal gas, meaning the internal energy does not increase with temperature for constant volume.
• (C) This is correct. The pressure of an ideal gas increases with temperature at constant volume, so .
• (D) This is incorrect. The volume of an ideal gas depends inversely on pressure for a given temperature, so .

Thus, the correct answers are (A), (C).

According to the kinetic theory of gases, the average kinetic energy of any ideal gas is directly proportional to the temperature and is independent of the type of gas. Since nitrogen and oxygen are both gases at the same temperature, their average kinetic energies are equal, as described by the equation:

Average kinetic energy

where is the Boltzmann constant and is the temperature. Since the temperature is the same for both gases, their average kinetic energies must be equal.

Thus, the correct answer is (D)