A gas mixture contains 75.2% nitrogen and 24.8% krypton by mass.

If the total pressure of the mixture is 745mmHg, calculate the partial pressure acting on the krypton in that given mixture.

This question aims to find the partial pressure exerted by an individual component of a gaseous mixture.

The basic concept behind this article on Dalton’s Law of Partial Pressure states that the total pressure that is exerted by a mixture of gases is the accumulative sum of individual pressures of individual gas elements that make up the mixture. It is represented as follows:

$P_{T o t a l} = P_{G a s 1} + P_{G a s 2} + P_{G a s 3} + \dots \dots$

It can also be expressed in terms of the number of moles or mole fraction:

$P_{G a s 1} = X_{G a s 1} {\times P}_{T o t a l}$

Here $X_{G a s 1}$ is the Mole Fraction for Gas 1 which is represented as follows in terms of number of moles $n$ :

$X_{G a s 1} = \frac{N u m b e r o f m o l e s o f G a s 1}{S u m o f N u m b e r o f m o l e s o f a l l G a s e s i n t h e m i x t u r e} = \frac{n_{G a s 1}}{n_{G a s 1} + n_{G a s 2} + n_{G a s 3} + \dots . .}$

Expert Answer

Given that:

Percentage of Nitrogen Gas in the gaseous mixture $N_{2} = 75.2$

Percentage of Krypton Gas in the gaseous mixture $K r = 24.8$

Total Pressure of the Gas Mixture $P_{T o t a l} = 745 m m H g$

Molar Mass of $N_{2} = 28.013 \frac{g}{m o l}$

Molar Mass of $K r = 83.798 \frac{g}{m o l}$

We know that percentage of a gaseous component in a gas mixture represents the mass of the individual gas in grams $g$ per $100 g$ of that particular gas mixture. Hence:

$75.2 % o f N_{2} = 75.2 g o f N_{2}$

$24.8 % o f K r = 24.8 g o f K r$

First, we will convert the given masses of individual gasses into the number of moles using molar mass.

We know that:

$N u m b e r o f M o l e s = \frac{G i v e n M a s s}{M o l a r M a s s}$

$n = \frac{m}{M}$

So, by using the above formula:

For Nitrogen Gas $N_{2}$ :

$n_{N_{2}} = \frac{75.2 g}{28.013 \frac{g}{m o l}}$

$n_{N_{2}} = 2.684 m o l$

For Krypton Gas $K r$ :

$n_{K r} = \frac{24.8 g}{83.798 \frac{g}{m o l}}$

$n_{K r} = 0.296 m o l$

Now we will use the Mole Fraction formula for Krypton Gas as follows:

$X_{K r} = \frac{n_{K r}}{n_{K r} + n_{N_{2}}}$

$X_{K r} = \frac{0.296 m o l}{0.296 m o l + 2.684 m o l}$

$X_{K r} = 0.0993$

To calculate the Partial Pressure of Krypton $K r$ , we will use Dalton’s Law of Partial Pressure in terms of Mole Fraction as follows:

$P_{K r} = X_{K r} {\times P}_{T o t a l}$

Substituting the given and calculated values in the above equation:

$P_{K r} = 0.0993 \times 745 m m H g$

$P a r t i a l P r e s s u r e o f K r y p t o n G a s P_{K r} = 74.0 m m H g$

Numerical Result

$24.8$ of Krypton Gas $(K r)$ in a gaseous mixture having a total pressure of $745 m m H g$ will exert an individual partial pressure of $74 m m H g$ .

$P a r t i a l P r e s s u r e o f K r y p t o n G a s P_{K r} = 74.0 m m H g$

Example

A gaseous mixture comprising oxygen $21$ and Nitrogen $79$ exerts a total pressure of $750 m m H g$ . Calculate the partial pressure exerted by Oxygen.

Solution

Percentage of Oxygen Gas in the gaseous mixture $O_{2} = 21$

Percentage of Nitrogen Gas in the gaseous mixture $N_{2} = 79$

Total Pressure of the Gas Mixture $P_{T o t a l} = 750 m m H g$

Molar Mass of $O_{2} = 32 \frac{g}{m o l}$

Molar Mass of $N_{2} = 28.013 \frac{g}{m o l}$

We know that:

$21 % o f O_{2} = 21 g o f N_{2}$

$79 % o f N_{2} = 79 g o f K r$

We will convert the given masses of individual gasses into the number of moles using molar mass.

For Oxygen Gas $O_{2}$ :

$n_{O_{2}} = \frac{21 g}{32 \frac{g}{m o l}}$

$n_{O_{2}} = 0.656 m o l$

For Nitrogen Gas $N_{2}$ :

$n_{N_{2}} = \frac{79 g}{28.013 \frac{g}{m o l}}$

$n_{N_{2}} = 2.82 m o l$

In order to calculate the Partial Pressure of Oxygen $O_{2}$ , we will use the Dalton’s Law of Partial Pressure in terms of Mole Fraction as follows:

$P_{O_{2}} = X_{O_{2}} {\times P}_{T o t a l}$

$P_{O_{2}} = \frac{n_{O_{2}}}{n_{N_{2}} + n_{O_{2}}} {\times P}_{T o t a l}$

$P_{O_{2}} = \frac{0.656 m o l}{0.656 m o l + 2.82 m o l} \times 750 m m H g$

$P a r t i a l P r e s s u r e o f O x y g e n G a s P_{O_{2}} = 141.54 m m H g$

Expert Answer

Numerical Result

Example

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