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 34.5kPa. In a seperate trial the student puts 214 mL of the same liquid into the same flask at the same conditions. What will the pressure measure (in kPa) in the second trial

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05-02-2021
Chemistry

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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 34.5kPa. In a seperate trial the student puts 214 mL of the same liquid into the same flask at the same conditions. What will the pressure measure (in kPa) in the second trial

Sagot :

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joetheelite joetheelite · Answer 1 · 2021-02-06T20:00:32-05:00

Answer:

See explanation below

Explanation:

In this case, we can use the Boyle's law. Assuming that the temperature of both trial remains constant, then:

P₁V₁ = P₂V₂ (1)

You should note that this expression is usable when you are dealing with gases. However, we can treat this unknown liquid as a gas, because all the air on the flask is removed, and we can assume that the liquid can behave like an ideal gas.

So using the above expression, we can solve for P₂:

P₂ = P₁V₁ / V₂ (2)

In this case, we already have the values of presures and volume, so replacing in this expression:

P₂ = 34.5 * 5 / 214

P₂ = 0.806 kPa

This should be the pressure of the liquid.

Hope this helps

Sagot :

P₂ = 0.806 kPa

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