A 1.00L balloon is filled with helium at 1.20 atm.If the balloon is sqeezed into a 0.500L beaker and doesnt burst ,what is the pressure of the helium?
p₁ =1.2 atm = 121590 Pa
V₁= 1 L=10⁻³ m³
V₂=0.5•10⁻³ m³
p₁V₁=p₂V₂
p₂=p₁V₁/V₂
To determine the pressure of helium in the 0.500L beaker after the balloon is squeezed, we need to apply the ideal gas law equation, which states:
PV = nRT
Where:
- P represents the pressure of the gas
- V represents the volume of the gas
- n represents the amount of substance (in moles)
- R is the ideal gas constant (0.0821 L·atm/(mol·K))
- T represents the temperature in Kelvin
To use this equation, we need to assume that the amount of helium inside the balloon remains constant when it is squeezed into the beaker. Therefore, the number of moles (n) of helium does not change.
Given:
- The initial volume of the balloon is 1.00L
- The initial pressure of helium inside the balloon is 1.20 atm
- The final volume of the balloon when squeezed into the beaker is 0.500L
- We need to find the final pressure of helium inside the beaker
Now let's set up the equation and solve for the final pressure (P):
(P1)(V1) = (P2)(V2)
Where:
- P1 = Initial pressure of helium inside the balloon = 1.20 atm
- V1 = Initial volume of the balloon = 1.00L
- P2 = Final pressure of helium inside the beaker (unknown)
- V2 = Final volume of the balloon when squeezed into the beaker = 0.500L
Plugging in these values into the equation:
(1.20 atm)(1.00 L) = (P2)(0.500 L)
Now, solving for P2:
P2 = (1.20 atm)(1.00 L) / (0.500 L)
P2 = 2.40 atm
Therefore, the pressure of helium inside the 0.500L beaker after the balloon is squeezed will be 2.40 atm.