Quiz Answers

1.

Estimate the thermal speed of an oxygen molecule at room temperature.

Hint:

There are N_A molecules in $1\,\mathrm{mol}$

Solution:

Since $\left\langle K\right\rangle=3kT/2$, average velocity is

$$\sqrt{\left\langle v^2\right\rangle} = \sqrt{2K/m}=\sqrt{3kT/m}$$

An average material scientist might not remember the mass of oxygen molecule, but she/he certainly remembers its molar mass $M=32\,\mathrm{g/mol}$. Then m=M/N_A. Recalling from Thermo course that $kN_A=R=8.3\,\mathrm{J/(K\cdot mol)}$, we obtain:

$$\sqrt{\left\langle v^2\right\rangle} = \sqrt{\frac{3RT}{M}} ... ...{3\cdot8\cdot300}{30\cdot10^{-3}}} \approx500\,\mathrm{m/s}$$

This is about $1800\,\mathrm{km/h}\approx 1000\,\mathrm{mile/h}$

2.

Calculate the chemical potential of ideal gas $\mu=(\partial A/\partial N)_{T,V}$.

Solution:

Since

$$A = -kTN\ln\frac{eV}{N} + Nf(T)$$

$$\begin{multline*} \mu = \left(\frac{\partial A}{\partial N}\right)_{T,V} = -kT\... ...T +f(T) =\ -kT\ln\frac{V}{N} +f(T) = kT\ln\frac{N}{V} +f(T) \end{multline*}$$
Chemical potential of ideal gas is kT log of the concentration (up to a constant)!

3.

Calculate the energy of the ideal gas without internal degrees of freedom ($q_{\text{internal}}=1$). If you are too lazy (or too smart) to calculate it from equation

 

E = Nf(T)-NTf'(T)

try to guess the answer. How would you do this?

Solution:

There are two ways to solve this problem.

(a)

Hard way. If $q_{\text{internal}}=1$,

$$f(T) = 3kT\ln\Lambda = \frac32 kT\ln\frac{2\pi\hbar}{mkT}$$

and

$$f'(T) = \frac32 k\ln\frac{2\pi\hbar}{mkT} - \frac32k$$

so

$$E = \frac32 NkT$$

(b)

Smart way. If there are no internal degrees of freedom, all the energy is kinetic energy K. This is 3kT/2 per particle, or 3NkT/2 per N particles