Question

Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at 0°C. Mass of a helium molecule = 6.64 × 10−27 kg and Boltzmann constant = 1.38 × 10−23 J K−1.

Answer 1

The average magnitude of linear momentum of a helium molecule in a sample of helium gas at 0°C is $8 \times 10^{-24}$.

Explanation:

Step 1:

Given data in the question  :

$m = 6.64 \times 10^{-27} kg$

T = 273 K

M is mass of a molecule of helium

T is temperature  

Step 2:

The average speed of Helium (He) atom is given by

$\begin{aligned}&V_{a v g}=\sqrt{\frac{8 R T}{\pi M}}\\&V_{a v g}=\sqrt{\frac{8 \times 8.3 \times 273}{3.14 \times 4 \times 10^{-3}}}\\&V_{a v g}=\sqrt{\frac{66.4 \times 273}{12.56 \times 10^{-3}}}\\&V_{a v g}=\sqrt{\frac{18127.2}{0.01256}}\end{aligned}$  

 $\begin{aligned}&V_{a v g}=\sqrt{1443248.408}\\&V_{a v g}=1201.35\end{aligned}$

Step 3:

We know that  

Momentum = m × Vavg

$\begin{aligned}&=6.64 \times 10^{-27} \times 1201.35\\&=7.97 \times 10^{-24}\\&=8 \times 10^{-24} \mathrm{kg}-\mathrm{m} / \mathrm{s}\end{aligned}$

   

Answer 2

The ideal gas law can be written in terms of the number of molecules of gas: PV = NkT, where P is pressure, V is volume, T is temperature, N is number of molecules, and k is the Boltzmann constant k = 1.38 × 10–23 J/K. A mole is the number of atoms in a 12-g sample of carbon-12.