Question

A) a scuba tank has a volume of 11.0L and is pressurized to 2975 PSI. assuming temperature of 25C, what would the volume of the gas be if it were all released from the cylinder at an ambient pressure of 758 Torr.

B) If the gas in the cylinder is 80% nitrogen and 20% oxygen, what is the mass of the gas in the cylinder.

Answer 1

Answer A-

Given,

Volume of Scuba Tank = 11.0 L

Pressure of Scuba tank = 2975 Psi or 202.4367 atm [1 Psi = 0.068046 atm]

Temperature = 25 C or 298 K

New Pressure = 758 Torr or 1 atm approx [1 atm = 760 torr]

New Volume = ?

We know that,

PV = nRT

where,

P = Pressure

V = Volume

n = moles

R = Gas Constant

T = temperature

For Initial Conditions -

PV = nRT

202.4367 atm * 11.0 L = nR * 298 K

(202.4367 atm*11.0 L)/298 K = nR ---------------A

Now for final condition,

PV = nRT

1 atm * V = nR * 298 K

(1 atm * V)/298 K = nR----------------------------B

Put A in B,

(1 atm * V)/298K = (202.4367 atm * 11.0 L)/298K

V = 2226.8 L [Answer]

Answer B -

Given,

80% Gas is Nitrogen

20% gas is Oxygen

Mass of gas in Cylinder = ?

We know that,

PV = nRT

where,

P = Pressure

V = Volume

n = moles

R = Gas Constant (0.08206 L⋅atm⋅K⁻¹⋅mol⁻¹)

T = temperature

For Initial Conditions -

PV = nRT

202.4367 atm * 11.0 L = n * 0.08206 L⋅atm⋅K⁻¹⋅mol⁻¹ * 298 K

n = (202.4367 atm*11.0 L)/(298 K * 0.08206 L⋅atm⋅K⁻¹⋅mol⁻¹)

n = 91.06 mol (Total moles of gas)

Now,

80% of Gas is Nitrogen and 20% is Oxygen.

So, Moles of Nitrogen = 80% * Total Moles

Moles of Nitrogen = 80% * 91.06 mol = 72.85 mol

Moles of Oxygen = 20% * Total Moles

Moles of Oxygen = 20% * 91.06 mol = 18.21 mol

Also,

Moles = mass/ Molar Mass

Mass = Moles * Molar mass

So,

Mass of 72.85 mol of Nitrogen =72.85 mol * 28.014 g = 2040.82 g

Mass of 18.21 mol of Oxygen = 18.21 mol * 31.9988 g/mol = 582.7 g

So, Total mass of gas = 2040.82 g + 582.7 g = 2623.52 g

MASS OF GAS IN CYLINDER IS 2623.52 g or 2.62 kg [ANSWER]