How is the partial pressure of a component gas proportional to the percentage composition (by molar quantity) to the gaseous mixture?

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How is the partial pressure of a component gas proportional to the percentage composition (by molar quantity) to the gaseous mixture?

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Answer 1 · 2017-02-14T19:43:07

The partial pressure of each gas is proportional to the given percentage.

Explanation:

$P_{Total} = P_{methane} + P_{ethane} + P_{propane} + P_{butane}$ $P_"Total"=P_"methane"+P_"ethane"+P_"propane"+P_"butane"$ .

Given the Ideal Gas equation, $P_{Total} = \frac{n_{Total} R T}{V}$ $P_"Total"=(n_"Total"RT)/V$

But $n_{Total} = (n_{methane} + n_{ethane} + n_{propane} + n_{butane})$ $n_"Total"=(n_"methane"+n_"ethane"+n_"propane"+n_"butane")$

And so,

$P_{Total} = (n_{methane} + n_{ethane} + n_{propane} + n_{butane}) \frac{R T}{V}$ $P_"Total"=(n_"methane"+n_"ethane"+n_"propane"+n_"butane")(RT)/V$

You have the percentage composition of the gas, and you simply multiply $P_{Total}$ $P_"Total"$ by the individual percentages to get the partial pressure of each gas.

Thus $P_{methane} = 1.48 \cdot a t m \times 94 % = ? ? \cdot a t m$ $P_"methane"=1.48*atmxx94%=??*atm$ . And you can work out these pressures individually.

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How is the partial pressure of a component gas proportional to the percentage composition (by molar quantity) to the gaseous mixture?

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