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 3.25×105 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 1.60×105 Pa .

Part A

Find the initial mass of oxygen. (in Kg)

(Following the steps submitted by another person for this problem, I got the mass = .301616 Kg and it was wrong)

Part B

Find the mass of oxygen that has leaked out. (in Kg)

If you are able, please include your steps so I can understand the math. Thank you for your help.

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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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)

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

hi there, i would say your answer is:

b. there is a fixed quantity of resources and technology available.

hope that : )