What pressure would be required to compress 7.75 litre of hydrogen at atmospheric pressure 5 litre (1.atom=760mmhg=760tors)
keeping temp constant..
P1/P2=V2/V1
or Pnew=1atm*7.75/5.00
To find the pressure required to compress 7.75 liters of hydrogen from an initial volume of 5 liters at atmospheric pressure (1 atm), we can use Boyle's Law.
Boyle's Law states that the pressure of a gas is inversely proportional to its volume, assuming the temperature remains constant. Mathematically, it can be expressed as:
P1 * V1 = P2 * V2
Where:
P1 = Initial pressure
V1 = Initial volume
P2 = Final pressure
V2 = Final volume
In this case, we know:
P1 = 1 atm
V1 = 5 liters
V2 = 7.75 liters
Let's rearrange the formula to solve for P2:
P2 = (P1 * V1) / V2
Substituting the given values:
P2 = (1 atm * 5 liters) / 7.75 liters
P2 = 5/7.75 atm
To convert the pressure from atm to mmHg or torr, we can use the conversion factor:
1 atm = 760 mmHg = 760 torr.
To find the pressure in mmHg:
P2 (in mmHg) = (5/7.75) * 760 mmHg
P2 (in mmHg) ≈ 489.67 mmHg
Therefore, the pressure required to compress 7.75 liters of hydrogen from 5 liters at atmospheric pressure is approximately 489.67 mmHg.