To solve the problem, we will follow these steps:
### Step 1: Convert the given values to appropriate units
- Volume \( V = 20 \, \text{L} = 20 \times 10^{-3} \, \text{m}^3 \)
- Temperature \( T = 27^\circ C = 27 + 273 = 300 \, \text{K} \)
- Pressure \( P = 2.0 \, \text{atm} = 2.0 \times 101325 \, \text{Pa} = 202650 \, \text{Pa} \)
### Step 2: Use the Ideal Gas Law to find the total number of moles \( n \)
The Ideal Gas Law is given by:
\[
PV = nRT
\]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
Substituting the values:
\[
n = \frac{PV}{RT} = \frac{(202650 \, \text{Pa})(20 \times 10^{-3} \, \text{m}^3)}{(8.314 \, \text{J/(mol K)})(300 \, \text{K})}
\]
Calculating \( n \):
\[
n = \frac{4053}{2494.2} \approx 1.62 \, \text{moles}
\]
### Step 3: Set up equations for the masses of hydrogen and helium
Let:
- \( m_H \) = mass of hydrogen
- \( m_{He} \) = mass of helium
We know:
\[
m_H + m_{He} = 5 \, \text{g} \quad \text{(1)}
\]
And the number of moles can be expressed as:
\[
\frac{m_H}{M_H} + \frac{m_{He}}{M_{He}} = n \quad \text{(2)}
\]
Where:
- \( M_H = 2 \, \text{g/mol} \) (molar mass of hydrogen)
- \( M_{He} = 4 \, \text{g/mol} \) (molar mass of helium)
Substituting the masses:
\[
\frac{m_H}{2} + \frac{m_{He}}{4} = 1.62 \quad \text{(2)}
\]
### Step 4: Solve the equations simultaneously
From equation (1):
\[
m_{He} = 5 - m_H
\]
Substituting \( m_{He} \) in equation (2):
\[
\frac{m_H}{2} + \frac{(5 - m_H)}{4} = 1.62
\]
Multiplying through by 4 to eliminate the fractions:
\[
2m_H + 5 - m_H = 6.48
\]
\[
m_H + 5 = 6.48
\]
\[
m_H = 6.48 - 5 = 1.48 \, \text{g}
\]
Now, substituting \( m_H \) back to find \( m_{He} \):
\[
m_{He} = 5 - 1.48 = 3.52 \, \text{g}
\]
### Step 5: Find the ratio of the mass of hydrogen to helium
The ratio \( \frac{m_H}{m_{He}} \) is:
\[
\frac{m_H}{m_{He}} = \frac{1.48}{3.52} = \frac{2}{5}
\]
### Final Answer
The ratio of the mass of hydrogen to helium in the given mixture is \( \frac{2}{5} \).
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