A 5 mL sample of an unknown liquid is placed into a 250 mL flask that has had all of the air removed. The pressure measured in the flask is found to be 39.9kPa. In a separate trial the student puts 214 mL of the same liquid into the same flask at the same temperature. What will the pressure measure (in kPa) in the second trial?

I did: (39.9kPa/5mL) * 214mL = 1707.72kPa
Is this correct?

In the first trial, was at least a drop of the liquid present when P was 39.9 kPa? If so then that is the vapor pressure of the liquid at that temperature and the pressure in the second trial will be 39.9 kPa.

Thank you. I just asked my tutor online for help, so I do not need any help anymore. Thank you anyways! :)

To calculate the pressure in the second trial, you can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when the temperature and amount of gas are held constant.

Boyle's Law can be mathematically expressed as P₁V₁ = P₂V₂, where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.

In the first trial, you have a 5 mL sample of liquid in a 250 mL flask, and the pressure measured is 39.9 kPa.

In the second trial, you have 214 mL of the same liquid in the same flask, and you need to calculate the pressure.

Using Boyle's Law, you can set up the equation as follows:
P₁V₁ = P₂V₂

Plugging in the values you have:
39.9 kPa * 250 mL = P₂ * 214 mL

To solve for P₂, you need to rearrange the equation:
P₂ = (39.9 kPa * 250 mL) / 214 mL

Calculating this expression:
P₂ = 46.63 kPa (rounded to two decimal places)

So, the pressure in the second trial is approximately 46.63 kPa, not 1707.72 kPa.