Question:
A welder using a tank of volume 7.50×10^-2 m³ fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.25×10^5 Pa and temperature of 37.9°C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 22.6°C, the gauge pressure of the oxygen in the tank is 1.60×10^5 Pa .
a. Find the initial mass of oxygen. (in Kg)
b. Find the mass of oxygen that has leaked out. (in Kg)
Answer:
a. The initial Mass of Oxygen is 301.76g
b. The mass of leaked oxygen is 145.6g
Explanation:
Given
Volume of Tank = V
V = 7.5 * 10^-2m³
V = 7.5/100m³
V = 0.075m³
Molar Mass = M
M = 32.0g/mol
Gauge Pressure 1 = P1
P1 = 3.25 * 10^5 Pa
Temperature 1 = T1
T1 = 37.9°C
T1 = 37.9 + 273K
T1 = 310.9K
Gauge Pressure 2 = P2
P2 = 1.60 * 10^5 Pa
Temperature 2 = T2
T2 = 22.6°C
T2 = 22.6 + 273K
T2 = 295.6K
a.
Initial Mass of Oxygen
Let m = Initial Mass before leakage
The relationship between m and M is
m = n * M
Where n is the number of moles of oxygen.
We'll calculate n using ideal gas equation
P * V = n * R * T ----- make n the subject of formula
n = PV/RT
Where P = P1, R = Universal Ɠas Constant = 8.315j/mol and T = T1
So,
n = ( 3.25 *10^5 * 0.075)/(8.315 * 310.9)
n = 24375/2585.1335
n = 9.43moles
So, m = n * M becomes
m = 9.43 * 32.0
m = 301.76g
b.
Mass of leaked gas
Leaked Gas = Initial Mass - Remaining Mass
Initial Mass = 301.76g
Calculating Remaining Mass...
The relationship between m and M is
m = n * M
Where m is the remaining mass
Where n is the number of moles of oxygen.
We'll calculate n using ideal gas equation
P * V = n * R * T ----- make n the subject of formula
n = PV/RT
Where P = P2, R = Universal Ɠas Constant = 8.315j/mol and T = T2
So,
n = (1.6 * 10^5 * 0.075)/(8.315 * 295.6)
n = 12000/2457.914
n = 4.88 moles
So, m = n * M becomes
m = 4.88 * 32
m = 156.16g
Leaked Gas = Initial Mass - Remaining Mass
Initial Mass = 301.76g
Remaining Mass = 156.16g
Leaked Gas = 301.76g - 156.16
Leaked Gas = 145.6g
The mass of oxygen that has leaked out is 141.12 g.
Using the equation;
PV = nRT
P =3.25×10^5 Pa or 3.2 atm
n = ?
R = 0.082 atmLK-1mol-1
T = 37.9 ∘C + 273 = 310.9 K
V = 7.50×10−2 m3 or 75 L
Now;
n = PV/RT
n = 3.2 atm × 75 L/0.082 atmLK-1mol-1 × 310.9 K
n = 9.4 moles
Mass of O2 = 9.4 moles × 32.0 g/mol = 300.8 g
On the other day;
P = 1.60×10^5 Pa or 1.6 atm
T = 22.6 ∘C + 273 = 295.6 K
V = 7.50×10−2 m3 or 75 L
n = PV/RT
n = 1.6 atm × 75 L/0.082 atmLK-1mol-1 × 295.6 K
n = 4.99 moles
Mass of O2 = 4.99 moles × 32.0 g/mol = 159.7 g
Mass of O2 that has leaked out = 300.8 g - 159.7 g = 141.12 g
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