A sample of helium at 20 °C occupies a volume of 3.63 L at a pressure of 3.07 atm. What volume does this helium sample occupy if the pressure is reduced to 2.43 atm while maintaining the temperature at 20 °C?
P1V1 = P2V2
To solve this problem, you can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when the temperature is constant.
Boyle's Law can be expressed as: P1 * V1 = P2 * V2
Where:
P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume
In this case, the initial pressure (P1) is 3.07 atm, the initial volume (V1) is 3.63 L, the final pressure (P2) is 2.43 atm, and the final volume (V2) is what we are trying to find.
Plugging in the values into the Boyle's Law equation, we have:
3.07 atm * 3.63 L = 2.43 atm * V2
Now we can solve for V2:
V2 = (3.07 atm * 3.63 L) / 2.43 atm
V2 = 4.59 L
Therefore, the helium sample will occupy a volume of 4.59 L when the pressure is reduced to 2.43 atm while maintaining the temperature at 20 °C.