a cylinder of helium has a volume of 8.0 L. The pressure of the gas is 388 lb in -2 at 25*C. What volume will the helium occupy at a standard atmospheric pressure (14.7 lb in -2) assuming there is no temperature change?
P1V1 = P2V2
211
211 WHAT? V2 = 211 liters?
yes 211 L? is that right?
yes.
To find the volume of helium at a standard atmospheric pressure, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to the pressure, assuming the temperature remains constant.
Boyle's Law equation: P1V1 = P2V2
Given:
P1 = 388 lb in^(-2) (initial pressure)
V1 = 8.0 L (initial volume)
P2 = 14.7 lb in^(-2) (final pressure)
V2 = ? (final volume)
We can rearrange the equation to solve for V2:
V2 = (P1 * V1) / P2
Now, let's substitute the given values into the equation:
V2 = (388 lb in^(-2) * 8.0 L) / 14.7 lb in^(-2)
Calculating this expression will give us the final volume, V2:
V2 = 210.034 L
Therefore, the volume of helium at standard atmospheric pressure (14.7 lb in^(-2)) assuming no temperature change would be approximately 210.034 L.