To solve the problem, we need to determine the volume of water molecules in 1 L of steam at a given temperature. We will use the concept of root mean square (r.m.s.) velocity and the ideal gas law.
### Step-by-Step Solution:
1. **Understanding r.m.s. Velocity**:
The r.m.s. velocity (C_rms) of a gas is given by the formula:
\[
C_{rms} = \sqrt{\frac{3RT}{M}}
\]
where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( M \) is the molar mass of the gas in kg/mol.
2. **Given Information**:
- The r.m.s. velocity of hydrogen (H₂) is \( \sqrt{7} \) times the r.m.s. velocity of nitrogen (N₂).
- The density of nitrogen is \( 1.0 \, \text{g/cm}^3 \) or \( 1000 \, \text{kg/m}^3 \).
- The density of water vapor is \( 0.0006 \, \text{g/cm}^3 \) or \( 0.6 \, \text{kg/m}^3 \).
3. **Calculating Molar Mass**:
- The molar mass of hydrogen (H₂) is \( 2 \, \text{g/mol} \) or \( 0.002 \, \text{kg/mol} \).
- The molar mass of nitrogen (N₂) is \( 28 \, \text{g/mol} \) or \( 0.028 \, \text{kg/mol} \).
4. **Setting Up the r.m.s. Velocity Equation**:
For hydrogen:
\[
C_{rms, H_2} = \sqrt{\frac{3RT}{0.002}}
\]
For nitrogen:
\[
C_{rms, N_2} = \sqrt{\frac{3RT}{0.028}}
\]
5. **Equating the r.m.s. Velocities**:
According to the problem, we have:
\[
C_{rms, H_2} = \sqrt{7} \cdot C_{rms, N_2}
\]
Substituting the expressions for r.m.s. velocities:
\[
\sqrt{\frac{3RT}{0.002}} = \sqrt{7} \cdot \sqrt{\frac{3RT}{0.028}}
\]
6. **Simplifying the Equation**:
Cancel \( \sqrt{3RT} \) from both sides:
\[
\frac{1}{\sqrt{0.002}} = \sqrt{7} \cdot \frac{1}{\sqrt{0.028}}
\]
Squaring both sides gives:
\[
\frac{1}{0.002} = 7 \cdot \frac{1}{0.028}
\]
Rearranging gives:
\[
0.028 = 14 \cdot 0.002
\]
7. **Finding the Volume of Water Molecules**:
Using the ideal gas law \( PV = nRT \), we can find the number of moles of water vapor in 1 L (or 0.001 m³) at standard temperature and pressure (STP).
\[
n = \frac{PV}{RT}
\]
Assuming \( P = 101325 \, \text{Pa} \) and \( R = 8.314 \, \text{J/(mol K)} \):
\[
n = \frac{(101325)(0.001)}{(8.314)(273.15)} \approx 0.0446 \, \text{mol}
\]
8. **Calculating the Volume of Water Molecules**:
The molar volume of water at STP is approximately \( 18 \, \text{g/mol} \). Therefore, the mass of water in 1 L of steam is:
\[
\text{mass} = n \cdot 18 \approx 0.0446 \cdot 18 \approx 0.803 \, \text{g}
\]
9. **Final Calculation**:
The volume of water molecules in 1 L of steam can be calculated using the density of water (1 g/cm³):
\[
\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{0.803 \, \text{g}}{1 \, \text{g/cm}^3} = 0.803 \, \text{cm}^3
\]
### Final Answer:
The volume of water molecules in 1 L of steam at this temperature is approximately **0.803 cm³**.
