Question

A scuba diver’s tank contains 0.29 kg of O2 compressed into a volume of 2.3L

a) What volume would this oxygen gas occupy at 26 °C and 96.3 kPa?

b) Using the definitions for density and Molecular Weight, what is the density 9 in g/L) of dioxygen gas at 77 °C and 700.0 torr?

Answer 1

a)

Molar mass of O2 = 32 g/mol


mass(O2)= 0.29 Kg
= 290.0 g

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(2.9*10^2 g)/(32 g/mol)
= 9.062 mol

Given:
P= 96.3 kPa
= (96.3*1000) pa
= 96300.0 Pa
= 96300.0 Pa
= (96300.0/101325) atm
= 0.9504 atm
n = 9.0625 mol
T = 26.0 oC
= (26.0+273) K
= 299 K

use:
P * V = n*R*T
0.9504 atm * V = 9.0625 mol* 0.08206 atm.L/mol.K * 299 K
V = 234 L
Answer: 234 L

b)
P= 700.0 torr
= (700.0/760) atm
= 0.9211 atm
T= 77.0 oC
= (77.0+273) K
= 350 K

Molar mass of O2 = 32 g/mol

Lets derive the equation to be used
use:
p*V=n*R*T
p*V=(mass/molar mass)*R*T
p*molar mass=(mass/V)*R*T
p*molar mass=density*R*T

Put Values:
0.9211 atm *32.0 g/mol = density * 0.08206 atm.L/mol.K *350.0 K
density = 1.0262 g/L
Answer: 1.03 g/L