The volume of a scuba tank is 10.0L.It containes a mixture of nitrogen and oxygen at 290.atm.what volume of this mixture could the tank supply to a diver at 2:40 atm?
V₂=T₂V₁/T₁
1 atm = 101325 Pa
1 L = 10⁻³ m³
To find the volume of the mixture that the tank could supply to the diver at 2.40 atm, we can use the combined gas law formula:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = initial pressure of the mixture (290 atm)
V1 = initial volume of the mixture (10.0 L)
T1 = initial temperature (assume constant for this calculation)
P2 = final pressure (2.40 atm)
V2 = final volume (what we want to find)
T2 = final temperature (assume constant for this calculation)
To solve for V2, rearrange the formula:
V2 = (P2 * V1 * T1) / (P1 * T2)
Since T1 and T2 are assumed to be constant and cancel each other out in this calculation, we can simplify the formula as:
V2 = (P2 * V1) / P1
Now let's plug in the values to get the volume:
V2 = (2.40 atm * 10.0 L) / 290 atm
V2 = 0.083 L (rounded to three decimal places)
Therefore, the volume of the mixture that the tank could supply to the diver at 2.40 atm is approximately 0.083 L.