
A sample of ${O_2}$ gas is collected over water at ${23^ \circ }C$ at a barometric pressure of $751\,mm\,Hg$ (vapor pressure of water at ${23^ \circ }C$ is $21\,mm\,Hg$). The partial pressure of ${O_2}$ gas in the sample collected is:
A. $21\,mm\,Hg$
B. $751\,mm\,Hg$
C. $0.96\,atm$
D. $1.02\,atm$
Answer
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Hint:Dalton’s law of partial pressure states that total pressure of a mixture of gases is equal to the sum of the partial pressure of the individual gases in the mixture. Ideally the ratio of partial pressures is equal to the ratio of numbers of molecules.
Complete answer:
Dalton’s law expresses the fact that the total pressure of the mixture of gases is equal to the sum of the partial pressure of gases present in that mixture. This law is applicable to ideal gas, as in ideal gas conditions gas molecules are far apart to even interact with each other.
$P = {P_{{N_2}}} + {P_{{H_2}}} + {P_{N{H_3}}}$.
Here P is total pressure, and ${P_{{N_2}}}$ is partial pressure of nitrogen gas, ${P_{{H_2}}}$ is partial pressure of hydrogen gas and ${P_{N{H_3}}}$ is partial pressure of ammonia gas in a container where ammonia gas is formed from nitrogen and hydrogen gas.
Here in the given question,
Total pressure is $751\,mm\,Hg$ at ${23^ \circ }C$
${x_i} = \dfrac{{{p_i}}}{p} = \dfrac{{{n_i}}}{n}$
Partial pressure of water given as $21\,mm\,Hg$ at ${23^ \circ }C$ and we have to calculate the partial pressure of oxygen gas. Applying Dalton’s law of partial pressure, we have
${P_{total}} = {P_{oxygen}} + {P_{water}}$
By putting the values, we get $751\,mm\,Hg = {P_{oxygen}} + 21\,mm\,Hg$
After calculation we get
$
{P_{oxygen}} = 751\,mm\,Hg - 21\,mm\,Hg \\
{P_{oxygen}} = 730\,mm\,Hg
$
By converting partial pressure of oxygen to atmospheric pressure, $\dfrac{{730\,mm\,Hg}}{{760\,mm\,Hg}} \times 1\,atm = 0.96\,atm$
Hence the correct option will be C.
Note:
The ratio of partial pressure equals the ratio of the number of molecules. That means, the mole fraction of an individual gas component in an ideal gas mixture can be expressed in terms of the component's partial pressure or moles of the component.
${x_i} = \dfrac{{{p_i}}}{p} = \dfrac{{{n_i}}}{n}$
Complete answer:
Dalton’s law expresses the fact that the total pressure of the mixture of gases is equal to the sum of the partial pressure of gases present in that mixture. This law is applicable to ideal gas, as in ideal gas conditions gas molecules are far apart to even interact with each other.
$P = {P_{{N_2}}} + {P_{{H_2}}} + {P_{N{H_3}}}$.
Here P is total pressure, and ${P_{{N_2}}}$ is partial pressure of nitrogen gas, ${P_{{H_2}}}$ is partial pressure of hydrogen gas and ${P_{N{H_3}}}$ is partial pressure of ammonia gas in a container where ammonia gas is formed from nitrogen and hydrogen gas.
Here in the given question,
Total pressure is $751\,mm\,Hg$ at ${23^ \circ }C$
${x_i} = \dfrac{{{p_i}}}{p} = \dfrac{{{n_i}}}{n}$
Partial pressure of water given as $21\,mm\,Hg$ at ${23^ \circ }C$ and we have to calculate the partial pressure of oxygen gas. Applying Dalton’s law of partial pressure, we have
${P_{total}} = {P_{oxygen}} + {P_{water}}$
By putting the values, we get $751\,mm\,Hg = {P_{oxygen}} + 21\,mm\,Hg$
After calculation we get
$
{P_{oxygen}} = 751\,mm\,Hg - 21\,mm\,Hg \\
{P_{oxygen}} = 730\,mm\,Hg
$
By converting partial pressure of oxygen to atmospheric pressure, $\dfrac{{730\,mm\,Hg}}{{760\,mm\,Hg}} \times 1\,atm = 0.96\,atm$
Hence the correct option will be C.
Note:
The ratio of partial pressure equals the ratio of the number of molecules. That means, the mole fraction of an individual gas component in an ideal gas mixture can be expressed in terms of the component's partial pressure or moles of the component.
${x_i} = \dfrac{{{p_i}}}{p} = \dfrac{{{n_i}}}{n}$
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