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# 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 $300\mathrm{K}$?

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Solution

## Step 1: Given informationMass 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 findWe have to determine how much oxygen leaks out.Step 3: CalculationWe 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,$\mathrm{m}\text{'}=\frac{\mathrm{P}\text{'}}{\mathrm{P}}×\frac{\mathrm{T}}{\mathrm{T}\text{'}}×\mathrm{m}$By substituting the given values, we get $=\frac{\frac{\mathrm{P}}{2}}{\mathrm{P}}×\frac{400}{300}×6\phantom{\rule{0ex}{0ex}}=4\mathrm{g}$Step 4: Calculating the mass of leaked oxygenCalculate the mass of oxygen leaked as follows:$⇒∆\mathrm{m}=\mathrm{m}-\mathrm{m}\text{'}\phantom{\rule{0ex}{0ex}}=6-4\phantom{\rule{0ex}{0ex}}=2\mathrm{g}$Therefore, the mass of oxygen leaked is $2\mathrm{g}$.

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