Question

A vessel has $6\mathrm{g}$ of oxygen at pressure P and temperature of $400\mathrm{K}$. A small hole is made in it so that oxygen leaks out. How much oxygen leaks out if the final pressure is at $\frac{\mathrm{P}}{2}$ and the temperature is $300\mathrm{K}$?

Answer 1

Step 1: Given information

Mass of oxygen is $6\mathrm{g}$ at temperature $400\mathrm{K}$.

The final pressure is at $\frac{\mathrm{P}}{2}$and the temperature is $300\mathrm{K}$.

Step 2: To find

We have to determine how much oxygen leaks out.

Step 3: Calculation

We know the relation,

$\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}}\mathrm{RT}...\left(1\right)$

$\mathrm{P}\text{'}\mathrm{V}=\frac{\mathrm{m}\text{'}}{\mathrm{M}}\mathrm{RT}\text{'}...\left(2\right)$

Where, $\text{P}$ is the pressure, $\mathrm{V}$ is the volume, $\mathrm{m}$ is the mass given in grams, $\mathrm{M}$ is the molar mass, $\mathrm{R}$ is the ideal gas constant, and $\mathrm{T}$ is the temperature in Kelvin.

By dividing $\left(1\right)$ and $\left(2\right)$, we get

$\frac{\mathrm{P}\text{'}}{\mathrm{P}}=\frac{\mathrm{m}\text{'}}{\mathrm{m}}\frac{\mathrm{T}\text{'}}{\mathrm{T}}$

Now,

$\mathrm{m}\text{'}=\frac{\mathrm{P}\text{'}}{\mathrm{P}}\times \frac{\mathrm{T}}{\mathrm{T}\text{'}}\times \mathrm{m}$

By substituting the given values, we get

$$\begin{gathered} =\frac{{\displaystyle \frac{\mathrm{P}}{2}}}{\mathrm{P}}\times \frac{400}{300}\times 6 \\ =4\mathrm{g} \end{gathered}$$

Step 4: Calculating the mass of leaked oxygen

Calculate the mass of oxygen leaked as follows:

$$\begin{gathered} \Rightarrow ∆\mathrm{m}=\mathrm{m}-\mathrm{m}\text{'} \\ =6-4 \\ =2\mathrm{g} \end{gathered}$$

Therefore, the mass of oxygen leaked is $2\mathrm{g}$.