What pressure would be required to compress 7.75 litres of hydrogen at 1atmospheric pressure to 5 litres?(1atm=760mm,Hg=760 torrant)
P
1
V
1
=
P
2
V
2
, now
P
1
=
1
⋅
a
t
m
, and so we solve for
P
2
:
P
2
=
P
1
V
1
V
2
=
1.00
⋅
L
×
1
⋅
a
t
m
0.050
⋅
L
=
?
?
a
t
m
This will be easier to read.
Use P1V1 = P2V2
P1 = 1 atm
V1 = 7.75 L
P2 = ?
V2 = 5 L
This assumes T does not change.
To calculate the pressure required to compress a gas, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when the temperature is constant. Mathematically, it can be written as:
P1 x V1 = P2 x V2
Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
In this case, we have:
P1 = 1 atm (given)
V1 = 7.75 litres (given)
V2 = 5 litres (given)
Now, we can rearrange the equation to solve for P2:
P2 = (P1 x V1) / V2
Substituting the given values:
P2 = (1 atm x 7.75 L) / 5 L
P2 = 7.75 atm
However, in the question, it is mentioned that 1 atmosphere is equal to 760 mmHg or 760 torr. So we need to convert 7.75 atm to mmHg or torr.
To convert atm to mmHg or torr, we can use the conversion factor:
1 atm = 760 mmHg = 760 torr
Converting:
P2 = 7.75 atm x 760 mmHg / 1 atm
P2 = 5894 mmHg
Therefore, the pressure required to compress 7.75 litres of hydrogen at 1 atmospheric pressure to 5 litres is approximately 5894 mmHg or 5894 torr.