Two moles of a monatomic gas with an rms speed of 294 m/s are contained in a tank that has a volume of 0.12 m³.If each gas particle has a mass of 8.220 × 10^-²⁶ kg,what is the absolute pressure of the gas?

Question

Quizplus · Accepted Answer

The absolute pressure of the gas can be calculated using the formula derived from the ideal gas law and the kinetic theory of gases: $P = \frac{2}{3} \cdot \frac{N}{V} \cdot \frac{1}{2}m(v_{rms})^2$, where $N$ is the number of molecules, $V$ is the volume, $m$ is the mass of a single molecule, and $v_{rms}$ is the root mean square speed of the molecules. Given $N = 2 \cdot N_A$ (where $N_A$ is Avogadro's number, since we have 2 moles), $V = 0.12 m^3$, $m = 8.220 \times 10^{-26} kg$, and $v_{rms} = 294 m/s$, substituting these values into the formula gives the pressure $P$, which after calculation results in approximately $2.4 \times 10^4 Pa$.

Two Moles of a Monatomic Gas with an Rms Speed