To analyze the scenario you've described, we'll refer to the ideal gas law which states:
\[ PV = nRT \]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles of gas,
- \( R \) is the ideal gas constant, and
- \( T \) is the temperature in Kelvin.
From your description, the initial volume \( V_1 \) is 10.0 nm (which I'll assume is a typo, as it is an improbably small volume and likely meant to be a larger unit), and the pressure fluctuates around a certain value \( P_1 \).
When the volume is decreased to half, \( V_2 = 5.0 nm \), the pressure will increase according to Boyle's Law, which states that for a fixed amount of gas at constant temperature, the pressure of a gas is inversely proportional to its volume:
\[ P_1 V_1 = P_2 V_2 \]
So, if the volume is halved:
\[ P_2 = \frac{P_1 V_1}{V_2} = \frac{P_1 \times 10.0}{5.0} = 2P_1 \]
This means the new pressure \( P_2 \) is around twice the initial pressure \( P_1 \).
In summary:
- Initial Pressure fluctuates around \( P_1 \).
- When the volume is decreased to 5.0 nm (assuming nm was a typo for standard units), the new pressure \( P_2 \) would be approximately \( 2P_1 \).