Question

Lisa needs to buy a helium tank to fill up all the balloons. If the helium must be compressed from an original volume of 27.0L at 14.7 psi, down to a volume of 3.00L, what would be the final pressure of the helium?

Answer 1

Answer:

$P_2=135.73psi$

Explanation:

Hello,

In this case we use the Boyle's law which allows us to understand the volume-pressure behavior as an inversely proportional relationship:

$V_1P_1=V_2P_2$

Whereas we solve for $P_2$ as the required final pressure:

$P_2=\frac{V_1P_1}{V_2} =\frac{27.0L*14.7psi}{3.00L}\\ \\P_2=135.73psi$

Best regards.

Answer 2

Answer:

The final pressure of the helium would be 132.2 psi or 9.0 atm

Explanation:

Step 1:data given

The original volume of the balloon = 27.0 L

The original pressure = 14.7 psi = 1.00 atm

The volume will be reduced to 3.00 L

Step 2: Calculate the pressure

P1*V1 = P2¨V2

⇒with P1 = The original pressure = 14.7 psi = 1.00 atm

⇒with V1 = The original volume of the balloon = 27.0 L

⇒with P2 = the new pressure = TO BE DETERMINED

⇒with V2 = the reduced volume = 3.00 L

14.7 psi * 27.0 L = P2 * 3.00 L

P2 = (14.7 psi * 27.0 L) / 3.00 L

P2 = 132.3 psi

OR

1 atm * 27.0 L = P2 * 3.00 L

P2 = (27.0 atm*L ) / 3.00 L

P2 = 9.0 atm

The final pressure of the helium would be 132.2 psi or 9.0 atm