You have 3 attempts to check you answer. Click the CHECK ANSWER button below. (2 points) In the virtual lab demo, a student adds particles to a container. While the temperature is held constant the initial volume of the container is 10.0 nm. The pressure fluctuates around what amount? When the container volume is changed to half its volume (5.0 nm), the pressure to the initial pressure amount. The new pressure is now about:

Question

GPT-4o mini · Answer

To determine the new pressure when the volume is halved while keeping the temperature constant, we can use Boyle's Law. Boyle's Law states that pressure and volume are inversely related in a closed system at constant temperature, which can be expressed as:

$P_{1} V_{1} = P_{2} V_{2}$

Where:

$P_{1}$ is the initial pressure,
$V_{1}$ is the initial volume,
$P_{2}$ is the new pressure,
$V_{2}$ is the new volume.

Given:

$V_{1} = 10.0, nm$
$V_{2} = 5.0, nm$

If we denote $P_{1}$ as the initial pressure (which is fluctuating around a certain value), the new pressure $P_{2}$ can be calculated as follows:

$P_{2} = P_{1} \times \frac{V_{1}}{V_{2}}$

Substituting the values:

$P_{2} = P_{1} \times \frac{10.0, nm}{5.0, nm} = P_{1} \times 2$

So, the new pressure $P_{2}$ will be roughly twice the initial pressure $P_{1}$ when the volume is halved to 5.0 nm.

If you need the numerical values, you would need to know the original fluctuating pressure value; however, the answer to your question is that the new pressure is approximately twice the initial pressure.