
In a gaseous mixture at $4$ atm pressure, \[25\% \] of molecules are Nitrogen, \[40\% \] of molecules are carbon dioxide and the rest are oxygen. The partial pressure of oxygen in the mixture is:
A. $1.40atm$
B. $1.6atm$
C. $1atm$
D. $0.9atm$
Answer
567.3k+ views
Hint: In the given question firstly we have to define the quantities of the significance that are given in the problem statement. Now after this we have to find the percentage of the oxygen. After that we can use the formula for the partial formula, \[\left( {partial{\text{ }}pressure} \right){O_2} = \;total{\text{ }}pressure\; \times \left( {mole{\text{ percentage}}} \right){O_2}\] . then we just have to apply the given values in the equation and find the answer.
Complete step by step answer:
Now firstly we have to sort out what are the given quantities in the given problem. They are :
The pressure at the gaseous mixture : $4$ atm
The percentage of the Nitrogen in the mixture : \[25\% \]
The percentage of the carbon dioxide in the mixture : \[40\% \]
The percentage of the oxygen in the mixture : $x\% $
So to find the percentage of the oxygen of the reaction, we just have to sum up others and then find difference from the total percentage. The equation would be :
$
100\% = 25\% + 40\% + x\% \\
100\% - 65\% = x\% \\
x\% = 35\% \\
$
Now after getting this we just have to get the formula for the partial pressure.
For the basic formulation of the partial pressure we get that the partial pressure equals :
${P_i} = Total{\text{ }}P \times {x_i}$
Where
${P_i}$ = partial pressure
$Total{\text{ }}P$ = total pressure
${x_i}$ = mole fraction
So the equation we need is the :
\[\left( {partial{\text{ }}pressure} \right){O_2} = \;total{\text{ }}pressure\; \times \left( {mole{\text{ percentage}}} \right){O_2}\]
Where we have to put all the values we have and solve to find the answer:
\[
\left( {partial{\text{ }}pressure} \right){O_2} = \;total{\text{ }}pressure\; \times \left( {mole{\text{ percentage}}} \right) \\
= 4 \times \dfrac{{35}}{{100}} \\
= \dfrac{7}{5} \\
= 1.40atm \\
\]
So, the correct answer is Option A.
Note: The partial pressure of a gas is a measure of thermodynamic activity of the gas's molecules. Gases dissolve, diffuse, and react according to their partial pressures, and not according to their concentrations in gas mixtures or liquids. This general property of gases is also true in chemical reactions of gases in biology.
Complete step by step answer:
Now firstly we have to sort out what are the given quantities in the given problem. They are :
The pressure at the gaseous mixture : $4$ atm
The percentage of the Nitrogen in the mixture : \[25\% \]
The percentage of the carbon dioxide in the mixture : \[40\% \]
The percentage of the oxygen in the mixture : $x\% $
So to find the percentage of the oxygen of the reaction, we just have to sum up others and then find difference from the total percentage. The equation would be :
$
100\% = 25\% + 40\% + x\% \\
100\% - 65\% = x\% \\
x\% = 35\% \\
$
Now after getting this we just have to get the formula for the partial pressure.
For the basic formulation of the partial pressure we get that the partial pressure equals :
${P_i} = Total{\text{ }}P \times {x_i}$
Where
${P_i}$ = partial pressure
$Total{\text{ }}P$ = total pressure
${x_i}$ = mole fraction
So the equation we need is the :
\[\left( {partial{\text{ }}pressure} \right){O_2} = \;total{\text{ }}pressure\; \times \left( {mole{\text{ percentage}}} \right){O_2}\]
Where we have to put all the values we have and solve to find the answer:
\[
\left( {partial{\text{ }}pressure} \right){O_2} = \;total{\text{ }}pressure\; \times \left( {mole{\text{ percentage}}} \right) \\
= 4 \times \dfrac{{35}}{{100}} \\
= \dfrac{7}{5} \\
= 1.40atm \\
\]
So, the correct answer is Option A.
Note: The partial pressure of a gas is a measure of thermodynamic activity of the gas's molecules. Gases dissolve, diffuse, and react according to their partial pressures, and not according to their concentrations in gas mixtures or liquids. This general property of gases is also true in chemical reactions of gases in biology.
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