A welder using a tank of volume 7.70×10 2 m3 fills it with oxygen (with a molar

ID: 1486689 • Letter: A

Question

A welder using a tank of volume 7.70×102 m3 fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.05×105 Pa and temperature of 39.0 C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 20.8 C, the gauge pressure of the oxygen in the tank is 1.90×105 Pa .

Part A

Find the initial mass of oxygen.

0.192

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Part B

Find the mass of oxygen that has leaked out.

0.096

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Part A

Find the initial mass of oxygen.

m =

0.192

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Part B

Find the mass of oxygen that has leaked out.

m =

0.096

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Explanation / Answer

V = 7.70x10^-2m^3

let us calculate the specific gas constant. R(oxygen) by dividing the universal gas constant 'R' by the molar mass 'M'.

R(oxygen) = R / M = 8.314 / 32.0 = 259.8J/Kg-K

HERE R = 8.314J/Kmole

using the eqn

PV = m RT---(1)

1) THE INITIAL MASS m1

p1 = 3.05x10^5Pa

T1 = 39 + 273 = 312K

m1 = p1 V / R T1

=(3.05x10^5)(7.70x10^-2) / (8.314)(312)

m1 = 9.05g-------(a)

2) Mass of oxygen that has leaked out is

m12 = m1 - m2------(2)

p2 = 1.90x10^5Pa

T2 = 20.8 + 273 = 293.8K

m2 = p2 V / RT2

=(1.90x10^5)(7.70x10^-2) / (8.314)(293.8)

m2 = 5.98g--------(b)

substituting the above values in eqn(2) we have

m12 = 9.05 - 5.98 = 3.07g

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A welder using a tank of volume 7.70×10 2 m3 fills it with oxygen (with a molar

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A welder using a tank of volume 7.70×10 2 m3 fills it with oxygen (with a molar

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