Question #bf6ac

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Answer 1 · 2014-12-09T12:13:31

The answer is $0.32 m m H g$ $0.32mmHg$ .

A simple way to approach the problem is by using the percentage of $C O_{2}$ $CO_2$ given, which equals the amount of $C O_{2}$ $CO_2$ present in the air sample, as well as its percentage of the total pressure.
According to Dalton's law of partial pressures, the total pressure measured for the sample is equal to

$P_{T O T A L} = P_{C O_{2}} + P_{r e s t}$ $P_(TOTAL) = P_(CO_2) + P_(rest)$ , where $P_{r e s t}$ $P_(rest$ represents the pressure the rest of the sample - that is, without the $C O_{2}$ $CO_2$ - would have if it occupied the same volume at the same temperature.

Since $C O_{2}$ $CO_2$ makes for $0.042 %$ $0.042%$ of the sample, we get

$P_{C O_{2}} = P_{T O T A L} \cdot \frac{0.042}{100} = 0.32 m m H g$ $P_(CO_2) = P_(TOTAL) * 0.042/100 = 0.32 mmHg$ , the prssure without the $C O_{2}$ $CO_2$ being $752 m m H g - 0.32 m m H g = 751.7 m m H g$ $752 mmHg - 0.32 mmHg = 751.7 mmHg$ .

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