Answer to Question #205151 in Chemistry for andrew

Question #205151


A sample of helium gas takes up 15 L. After being compressed down to a volume of 8 L, the sample has a pressure of 4.3 atm. What was the original pressure? 


1
Expert's answer
2021-06-10T03:34:54-0400

P1 = unknown

V1 = 15 L

P2 = 4.3 atm

V2 = 8 L

T1 = T2 = const


Solution:

Since the temperature and amount of gas remain unchanged, Boyle's law can be used.

Boyle's gas law can be expressed as: P1V1 = P2V2

To find the original pressure, solve the equation for P1:

P1 = P2V2 / V1

P1 = (4.3 atm × 8 L) / (15 L) = 2.29 atm = 2.3 atm

P1 = 2.3 atm


Answer: 2.3 atm is the original pressure.

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