What pressure would be required to compress 7.75 liters of hydrogen at atmospheric pressure to 5 liters
since PV is constant, you want P such that
7.75 * 1 = 5P
To determine the pressure required to compress 7.75 liters of hydrogen to 5 liters at atmospheric pressure, we can use Boyle's Law, which states that the product of the initial pressure and volume is equal to the product of the final pressure and volume for a given amount of gas, assuming constant temperature.
Boyle's Law equation: P1 * V1 = P2 * V2
Given:
V1 = 7.75 liters (initial volume)
V2 = 5 liters (final volume)
P1 = atmospheric pressure (let's assume it's 1 atmosphere)
P2 = unknown
Substituting the given values into the equation, we have:
1 atmosphere * 7.75 liters = P2 * 5 liters
Simplifying the equation:
7.75 = 5P2
Now, solve for P2 by dividing both sides of the equation by 5:
P2 = 7.75 / 5
Calculating the value:
P2 = 1.55 atmospheres
Therefore, to compress 7.75 liters of hydrogen at atmospheric pressure to 5 liters, a pressure of 1.55 atmospheres would be required.