To determine how the pressure changes when the volume of the container is altered, we can use the ideal gas law, expressed as:
\[ PV = nRT \]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the amount of substance (number of particles),
- \( R \) is the ideal gas constant, and
- \( T \) is the temperature.
Assuming the number of particles \((n)\) and temperature \((T)\) remain constant, when you halve the volume (from 10.0 nm to 5.0 nm), we can analyze the relationship between pressure and volume. According to Boyle's law, pressure is inversely proportional to volume when temperature and the amount of gas are held constant.
If the initial volume \( V_1 = 10.0 , \text{nm} \) and the initial pressure \( P_1 \) is approximately around some value (let's say \( P_1 \)) and you change the volume to \( V_2 = 5.0 , \text{nm} \), the relationship can be expressed as:
\[ P_1 V_1 = P_2 V_2 \]
Now substituting the volumes gives us:
\[ P_1 \cdot 10.0 , \text{nm} = P_2 \cdot 5.0 , \text{nm} \]
Solving for \( P_2 \):
\[ P_2 = \frac{P_1 \cdot 10.0 , \text{nm}}{5.0 , \text{nm}} = 2P_1 \]
This means that the new pressure \( P_2 \) is approximately twice the initial pressure \( P_1 \).
In summary, if the initial pressure fluctuated around a value \( P_1 \), upon reducing the volume to 5.0 nm, the new pressure would be about \( 2P_1 \), meaning the pressure would now be approximately doubled.