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A 2. 26 l balloon of helium is at 30°c and 1. 61 atm, what is the pressure if the volume is increased to 4. 12 l?.

User Olala
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1 Answer

24 votes
24 votes

Answer:

0.8832 atm

Step-by-step explanation:

Per Boyle's Law:

If the temperature is held constant, the volume of a fixed amount of gas is inversely proportional to its pressure:

V∝
(1)/(P)

and


P_1V_1=P_2V_2

To find pressure resulting from a change in volume:


(P_1V_1)/(V_2)=P_2

So:


P_1=1.61atm\\V_1=2.26L\\V_2=4.12L\\P_2=((1.61atm)(2.26L))/(4.12L)\\P_2=0.8832atm

This answer makes sense because Boyle's Law states that a given volume of gas at a constant temperature should have a proportionally lower pressure as the volume is increased.

User Murthi
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