Chapter 5, Problem 5.152P

Interpretation Introduction

Interpretation:

The ratio of $\frac{{\text{O}}_{\text{2}}}{{\text{H}}_{\text{2}}}$ in the balloon the next day is to be calculated.

Concept introduction:

Effusion is explained as the movement of the gas molecule through a pinhole.

Diffusion can be explained as the mixing of one gas molecule with another gas molecule by random motion.

According to Graham’s law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The mathematical expression of Graham’s law of effusion is as follows:

$\frac{\text{Rate A}}{\text{Rate B}}=\frac{u_{rms\text{ of A}}}{u_{rms\text{ of B}}}=\frac{\sqrt{M_{\text{B}}}}{\sqrt{M_{\text{A}}}}=\frac{\text{time B}}{\text{time A}}=\frac{\text{Moles of A}}{\text{Moles of B}}$

Here, $M_{\text{B}}$ is the molar mass of gas $\text{B}$.

$M_{\text{A}}$ is the molar mass of gas $\text{A}$.

The ideal gas equation can be expressed as follows,

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

The expression to calculate the partial pressure of the gas is as follows:

$P_{\text{gas}}=X_{\text{gas}}\cdot P_{\text{total}}$

Here, $P_{\text{gas}}$ is the partial pressure of the gas, $X_{\text{gas}}$ mole fraction of the gas and $P_{\text{total}}$ is the total pressure.

The expression to calculate the mole fraction of the gas is as follows,

$X_{\text{gas}}=\frac{\text{Moles of gas}}{\text{Total moles}}$

Here, $X_{\text{gas}}$ is the mole fraction of the gas.