Problem — True or false on ideal gas and speed

Problem

State whether the following statements are true or false, briefly justifying your answer:

1. In an ideal gas the molecules interact through small attractive forces.
2. At equal temperature, a monatomic gas and a diatomic gas have the same root-mean-square speed for a molecule of equal mass.
3. The pressure of a gas depends only on temperature.
4. The factor $3$ in $v_\text{rms}$ comes from the three translational degrees of freedom.

Solution

1. FALSE. In the ideal-gas model the molecules are point-like and do not interact at a distance: no attractive or repulsive forces, only instantaneous elastic collisions. Attractive forces appear in real gases (the van der Waals model).

2. TRUE. The root-mean-square speed depends only on the mass of the molecule and on temperature: $v_\text{rms} = \sqrt{\frac{3k_B T}{m}}$ The number of degrees of freedom does not appear: at equal $T$ and $m$, the value is identical for a monatomic and a diatomic molecule. The difference between the two lies in the total internal energy (which includes rotation), not in the translational motion.

3. FALSE. From the equation of state $P = \dfrac{nRT}{V}$: the pressure also depends on the number of moles and on the volume. At equal $T$, compressing the gas or adding moles changes the pressure.

4. TRUE. The mean translational kinetic energy is $\tfrac{3}{2}k_B T$, with a contribution of $\tfrac{1}{2}k_B T$ for each of the three translational degrees of freedom ($x,y,z$). From $\tfrac{1}{2}m\langle v^2\rangle = \tfrac{3}{2}k_B T$ the factor $3$ under the square root arises. $\ev{\text{F},\ \text{T},\ \text{F},\ \text{T}}$

Links

Topics: Kinetic theory of gases Concepts: Root-mean-square speed · Pressure Skills: Micro-macro interpretation