Question

What is the total translational KE of the atoms if the balloon has a diameter 39.0 cm at 46.0 °C and the pressure inside the balloon is 121.6 kPa?

Answer 1

What is the total translational KE of the atoms if the balloon has a diameter 39.0 cm at 46.0 °C and the pressure inside the balloon is 121.6 kPa?

The average kinetic energy of a gas molecule is

$$K E _ {1} = \frac {3}{2} k _ {B} T$$

where

$k_{B} = 1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$ - Boltzmann's constant;

$T = 273.15 + 46.0 = 319.15K$ - absolute temperature

$$K E = \frac {3}{2} * 1.38 \times 10^{-23} * 319.15 = 6.61 \times 10^{-21} \mathrm{~J~per~atom}$$

The number of atoms can be found from ideal gas law

$$\nu = \frac {P V}{R T} \text{ (moles)}$$

where $P$ is the pressure, $V$ is the volume, $\nu$ is the number of molecules present, $R$ is the gas constant (8.31J/(mol*K)), and $T$ is the temperature in Kelvins.

$$P = 121.6 \mathrm{~kPa}$$ $$V = \frac {4}{3} \pi \left(\frac {39}{2}\right) ^ {3} \times 10^{-6} \mathrm{~m}^{3}$$ $$\nu = \frac {121.6 \times 10^{3} \mathrm{~Pa} * \frac {4}{3} \pi \left(\frac {39}{2}\right) ^ {3} \times 10^{-6} \mathrm{~m}^{3}}{8.31 \mathrm{~J} \mathrm{~mol} * \mathrm{K} * 319.15 \mathrm{~K}} = 1.424 \mathrm{~mol}$$

Number of atoms

$$N = \nu N _ {A} = 1.424 * 6.022 \times 10^{23} = 8.58 \times 10^{23}$$

Total translational KE:

$$K E = N * K _ {1} = 8.58 \times 10^{23} * 6.61 \times 10^{-21} \mathrm{~J} = 5668.55 \mathrm{~J} \approx 5.67 \mathrm{~kJ}$$