Question

Suppose you double the temperature of a gas at constant volume. Do the following change? If so, by what factor?

a. The average translational kinetic energy of a mole cule.

b. The rms speed of a molecule.

c. The mean free path.

Short answer

(a) increase to double the initial value,

(b) increase to $\sqrt{2}$times the initial value,

(c) will not change if final pressure is doubled, twice if the final pressure is the same, otherwise cannot say.

Step-by-step solution

Step :1   kinetic energy (part a)

(a) TThe equation is as follows for the average translational energy and temperature relationship:

${ϵ}_{avg}=\frac{3}{2}{k}_{B}T$

Since everything apart from the temperature is a constant.

Step :2  Speed (part b)

(b) The rms speed is given as

$v_{rms}=\sqrt{\frac{3k_{B}T}{m}}=C\sqrt{T}$

where we have collected all the constants under $c$. This result means that the rms speed is proportional to the square root of the temperature. That is to say,

if the temperature is doubled, the rms speed will increase by the square root of two.

Step :3  Formula of mean (part c)

The formula of mean free path is

$\lambda =\frac{1}{4\sqrt{2}\pi N/V{r}^2}$

At first glance, the mean free path appears to be temperature agnostic. The numerical density, on the other hand, is temperature-dependent..

From the optmial gas law we have

$pV=N{k}_{B}T\Rightarrow \frac{N}{V}=\frac{p}{k_{B}T}$

The numerical density will fall by a factor of two if the temperature is doubled while the volume and pressure remain constant. However, until we know the end pressure, we can't say anything about the numerical density if the heating isn't also isobaric.