Question:
Equal masses of helium and oxygen are mixed in an empty container at $25^{\circ}\text{C}$. What is the fraction of the total pressure exerted by helium?
Equal masses of helium and oxygen are mixed in an empty container at $25^{\circ}\text{C}$. What is the fraction of the total pressure exerted by helium?
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Even though the masses are exactly the same, Helium exerts 8 times more pressure than Oxygen! Why? Because Helium atoms are 8 times lighter ($32/4 = 8$), meaning an equal mass contains 8 times more independent, pressure-exerting particles!
Updated On: Aug 19, 2026
- $1/2$
- $1/4$
- $8/9$
- $7/9$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We have a gaseous mixture containing physically equal masses of helium ($\text{He}$) and oxygen ($\text{O}_2$). We need to determine the ratio of the partial pressure of helium to the total pressure.
Step 2: Detailed Explanation:
According to Dalton's Law of Partial Pressures, the partial pressure of any gas in a mixture is directly proportional to its mole fraction.
$P_{\text{He}} = X_{\text{He}} \times P_{\text{total}}$
Therefore, the fraction of the total pressure exerted by helium is exactly equal to its mole fraction ($X_{\text{He}}$).
Let the equal mass of each gas added to the container be $m$ grams.
1. Calculate the number of moles of each gas:
Molar mass of Helium ($\text{He}$) = $4 \text{ g/mol}$
Moles of Helium ($n_{\text{He}}$) = $\frac{m}{4}$
Molar mass of Oxygen ($\text{O}_2$) = $32 \text{ g/mol}$
Moles of Oxygen ($n_{\text{O}_2}$) = $\frac{m}{32}$
2. Calculate the total number of moles:
$n_{\text{total}} = n_{\text{He}} + n_{\text{O}_2}$
$n_{\text{total}} = \frac{m}{4} + \frac{m}{32}$
Find a common denominator (32):
$n_{\text{total}} = \frac{8m}{32} + \frac{1m}{32} = \frac{9m}{32}$
3. Calculate the mole fraction of Helium ($X_{\text{He}}$):
$X_{\text{He}} = \frac{n_{\text{He}}}{n_{\text{total}}}$
$X_{\text{He}} = \frac{\frac{m}{4}}{\frac{9m}{32}}$
Multiply by the reciprocal:
$X_{\text{He}} = \frac{m}{4} \times \frac{32}{9m}$
The mass '$m$' cancels out perfectly:
$X_{\text{He}} = \frac{32}{36}$
$X_{\text{He}} = \frac{8}{9}$
Because the partial pressure fraction equals the mole fraction, Helium exerts $8/9$ of the total pressure.
Step 3: Final Answer:
The fraction of the total pressure is $8/9$, matching option (c).
We have a gaseous mixture containing physically equal masses of helium ($\text{He}$) and oxygen ($\text{O}_2$). We need to determine the ratio of the partial pressure of helium to the total pressure.
Step 2: Detailed Explanation:
According to Dalton's Law of Partial Pressures, the partial pressure of any gas in a mixture is directly proportional to its mole fraction.
$P_{\text{He}} = X_{\text{He}} \times P_{\text{total}}$
Therefore, the fraction of the total pressure exerted by helium is exactly equal to its mole fraction ($X_{\text{He}}$).
Let the equal mass of each gas added to the container be $m$ grams.
1. Calculate the number of moles of each gas:
Molar mass of Helium ($\text{He}$) = $4 \text{ g/mol}$
Moles of Helium ($n_{\text{He}}$) = $\frac{m}{4}$
Molar mass of Oxygen ($\text{O}_2$) = $32 \text{ g/mol}$
Moles of Oxygen ($n_{\text{O}_2}$) = $\frac{m}{32}$
2. Calculate the total number of moles:
$n_{\text{total}} = n_{\text{He}} + n_{\text{O}_2}$
$n_{\text{total}} = \frac{m}{4} + \frac{m}{32}$
Find a common denominator (32):
$n_{\text{total}} = \frac{8m}{32} + \frac{1m}{32} = \frac{9m}{32}$
3. Calculate the mole fraction of Helium ($X_{\text{He}}$):
$X_{\text{He}} = \frac{n_{\text{He}}}{n_{\text{total}}}$
$X_{\text{He}} = \frac{\frac{m}{4}}{\frac{9m}{32}}$
Multiply by the reciprocal:
$X_{\text{He}} = \frac{m}{4} \times \frac{32}{9m}$
The mass '$m$' cancels out perfectly:
$X_{\text{He}} = \frac{32}{36}$
$X_{\text{He}} = \frac{8}{9}$
Because the partial pressure fraction equals the mole fraction, Helium exerts $8/9$ of the total pressure.
Step 3: Final Answer:
The fraction of the total pressure is $8/9$, matching option (c).
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