2019-Nov-Physics_paper_2__TZ0_HL Detailed Solution

Solution:

The average speed of the helium atoms in the container can be calculated using the root-mean-square (rms) speed formula for ideal gases:

$$
v_{\text{rms}} = \sqrt{\frac{3kT}{m}}
$$

where $k$ is the Boltzmann constant, $T$ is the temperature in kelvins, $m$ is the mass of one helium atom, and $v_{\text{rms}}$ is the root-mean-square speed of the gas particles.

First, we need to find the number of helium atoms in the container. We can use the ideal gas law:

$$
PV = nRT
$$

where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of gas, $R$ is the universal gas constant, and $T$ is the temperature in kelvins.

Solving for $n$ gives:

$$
n = \frac{PV}{RT}
$$

Substituting the given values, we get:

$$
n = \frac{(5.1 \times 10^5~\mathrm{Pa})(3.2 \times 10^{-6}~\mathrm{m}^3)}{(8.31~\mathrm{J/(mol~K)})(320~\mathrm{K})} \approx 6.63 \times 10^{-8}~\mathrm{mol}
$$

Next, we can find the number of helium atoms in the container using Avogadro’s number:

$$
N = nN_{\text{A}}
$$

where $N_{\text{A}}$ is Avogadro’s number ($6.02 \times 10^{23}~\mathrm{mol}^{-1}$). Substituting the values, we get:

$$
N = (6.63 \times 10^{-8}~\mathrm{mol})(6.02 \times 10^{23}~\mathrm{mol}^{-1}) \approx 3.99 \times 10^{16}
$$

Finally, we can calculate the rms speed of the helium atoms:

$$
v_{\text{rms}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3(1.38 \times 10^{-23}~\mathrm{J/K})(320~\mathrm{K})}{6.6 \times 10^{-27}~\mathrm{kg}}} \approx 1400~\mathrm{m/s}
$$

Therefore, the average speed of the helium atoms in the container is approximately $1400~\mathrm{m/s}$.