What pressure would be required to compress 7.75litres of hydrogen at atmospheric pressure to 5litres?(1 atom=760mmhg=760 torr)
same temp?
PV=nRT, so if nRT is constant, PV=constant (Boyle's law).
If V goes from 7.5 to 5 (v2/v1= 5/7.5), pressure goes from 1atm to 5/7.5 atm.
P= consant/V
help please? i dont even know how to start
I need your help please
what pressure would be required to compressed 7.75 litres of hydrogen atmosphic pressure to 5 litres ( 1 atm=760 mmhg =760) torrent
Which kind of help
I don't know how to solve it
To determine the pressure required to compress hydrogen gas from 7.75 liters to 5 liters, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.
Boyle's Law equation: P₁V₁ = P₂V₂
Where:
P₁ = initial pressure (atmospheric pressure)
V₁ = initial volume (7.75 liters)
P₂ = final pressure (what we need to find)
V₂ = final volume (5 liters)
Given that atmospheric pressure is equivalent to 760 mmHg (or 760 torr), we can now solve for P₂:
P₁V₁ = P₂V₂
(1 atm) (7.75 L) = P₂ (5 L)
To convert atmospheric pressure to torr, we use the conversion factor:
1 atm = 760 torr
Now we can substitute the values:
(760 torr) (7.75 L) = P₂ (5 L)
By rearranging the equation and solving for P₂:
P₂ = (760 torr) (7.75 L) / (5 L)
P₂ ≈ 1185.6 torr
Therefore, approximately 1185.6 torr of pressure would be required to compress 7.75 liters of hydrogen at atmospheric pressure to 5 liters.