Question

HELP!

What mass, in grams, of oxygen gas (O2) is contained in a 10.5-liter tank at 27.3 degrees Celsius and 1.83 atmospheres? Show all of the work used to solve this problem.

Answer 1

Answer : The mass of oxygen gas contained in the tank is 24.9 grams.

Explanation :

Using ideal gas equation:

`PV=nRT\\\\PV=(w)/(M)RT`

where,

P = pressure of oxygen gas = 1.83 atm

T = temperature of oxygen gas =
`27.3^oC=273+27.3=300.3K`

V = volume of oxygen gas = 10.5 L

M = molar mass of oxygen gas = 32 g/mole

w = mass of oxygen gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the above equation, we get:

`PV=(w)/(M)RT`

`(1.83atm)* (10.5L)=(w)/(32g/mole)* (0.0821L.atm/mol.K)* (300.3K)`

`w=24.9g`

Therefore, the mass of oxygen gas contained in the tank is 24.9 grams.

Answer 2

Answer : The mass of oxygen gas is, 24.94 grams

Explanation :

Using ideal gas equation :

`PV=nRT`

or,

`PV=(w)/(M)RT`

where,

P = pressure of gas = 1.83 atm

V = volume of gas = 10.5 L

T = temperature of gas =
`27.3^oC=273+27.3=300.3K`

R = gas constant = 0.0821 L.atm/mole.K

n = number of moles of oxygen gas

w = mass of oxygen gas = ?

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above law, we get

`1.83atm* 10.5L=(w)/(32g/mole)* 0.0821L.atm/mole.K* 300.3K`

`w=24.94g`

Therefore, the mass of oxygen gas is, 24.94 grams