A 5.00L sample of helium expands to 12.0L at which point the pressure is measured to be 0.720 atm what is the original pressure of the gas ?

Question

asked Sep 17, 2021 143k views

2 Answers

Erikvold · Answer 1 · 2021-09-24T01:30:29+0000

Answer:

1.73 atm

Step-by-step explanation:

Given data:

Volume = V1 = 5L

Volume = V2 = 12.0 L

Pressure P2 = 0.720 atm

To fnd:

Pressure P1 = ?

Formula:

By using Boyle's law,

P1 x V1 = P2 x V2

Calculation:

P1 = (P2 x V2) / V1

= 0.720 atm X 12.0 L / 5L

= 8.64/5

= 1.73 atm