A mixture of gases in a cylinder at 760 $m m$ pressure contains 65% nitrogen, 15% oxygen, and 20% carbon dioxide by volume. What is the partial pressure of each gas in $m m$ ?

P_{N_{2}} = 494 m m; P_{O_{2}} = 114 m m; P_{C O_{2}} = 152 m m

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P_{N_{2}} = 114 m m; P_{O_{2}} = 494 m m; P_{C O_{2}} = 152 m m

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P_{N_{2}} = 152 m m; P_{O_{2}} = 114 m m; P_{C O_{2}} = 494 m m

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P_{N_{2}} = 494 m m; P_{O_{2}} = 152 m m; P_{C O_{2}} = 114 m m

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Solution

The correct option is A $P_{N_{2}} = 494 m m; P_{O_{2}} = 114 m m; P_{C O_{2}} = 152 m m$
$P_{N_{2}} = \frac{760 \times 65}{100} = 494 m m$
$P_{O_{2}} = \frac{760 \times 15}{100} = 114 m m$
$P_{C O_{2}} = \frac{760 \times 20}{100} = 152 m m$

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A mixture of gases in a cylinder at 760 mm pressure contains 65% nitrogen, 15% oxygen, and 20% carbon dioxide by volume. What is the partial pressure of each gas in mm?

The correct option is A PN2=494mm;PO2=114mm;PCO2=152mmPN2=760×65100=494mmPO2=760×15100=114mmPCO2=760×20100=152mm

A mixture of gases in a cylinder at 760 $m m$ pressure contains 65% nitrogen, 15% oxygen, and 20% carbon dioxide by volume. What is the partial pressure of each gas in $m m$ ?

The correct option is A $P_{N_{2}} = 494 m m; P_{O_{2}} = 114 m m; P_{C O_{2}} = 152 m m$
$P_{N_{2}} = \frac{760 \times 65}{100} = 494 m m$
$P_{O_{2}} = \frac{760 \times 15}{100} = 114 m m$
$P_{C O_{2}} = \frac{760 \times 20}{100} = 152 m m$