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

Question

A welder using a tank of volume 7.70×102m3 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.

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Part B

Find the mass of oxygen that has leaked out.

92.09

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A welder using a tank of volume 7.70×102m3 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.

m = kg

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Part B

Find the mass of oxygen that has leaked out.

m =

92.09

  kg

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Explanation / Answer

a)

pV = nRT

n = pv/RT

Given Guage pressure is 3.05×105 Pa

Absolute pressure is 3.05×105 Pa - 1.01325*105 Pa = 2.03675*105 Pa

n = pv/RT = 2.03675*105 *7.70×102/ (8.314*312 K) = 6

m1 = Mn; M = Molar Mass

m1 = 0.032 Kg/mol * 6 = 0.192 Kg

b)

the mass of oxygen that has leakes out 'm12'
m12 = m1 - m2
p2 =( 1.95 - 1.01325)×10^5Pa = 0.3675*10^5Pa
T2 = 20.8C = 293.8K

n2 = p2 * V / R * T2 =0.3675*10^5Pa*7.70×102m3 /(8.314*293.8) = 3

m2 = 0.032 Kg/mol * 3 = 0.096 Kg

m12 = m1 - m2 = 0.096 kg

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