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

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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.

0.192

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Part B

Find the mass of oxygen that has leaked out.

0.096

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p*V = n*R*T
(p=pressure, V=volume, m=mass,R= gas constant, T=temperature)

now n= number of moles= mass /Molecular mass

p*V =( m/M)*R*T

m=M P V/RT

V = 7.7×10^-2m^3

p1 = 3.05×10^5Pa

T1 = 39C = 39 + 273 = 312K

so m= (3.05×10^5) (7.7×10^-2)(0.032) /(8.314*312)

m=0.289kg = 289 grams

(b)let new mass = m1

m1=M PV/RT

T=20.8 o C = 20.8+ 273 =293.8K

m1=(0.032) (1.9x10^5)(7.7x10^-2)/(8.314 * 293.8)

m1= 0.192 kg = 192 g

so mass of oxygen that leaked out= m-m1 = 289-192 =97grams

this amser is not correct !!!!

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 =

0.192

kg

SubmitMy AnswersGive Up

Incorrect; Try Again; 5 attempts remaining

Part B

Find the mass of oxygen that has leaked out.

m =

0.096

  kg

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Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

First you have to calculate the specific gas constant RO2 by dividing the universtal gas constant 'R' by the molar mass 'M'
RO2 = R / M = (8.3144 J / K*mol) / (32.0g/mol) = 0.2598 J/g*K = 259.8 J/kg*K

p*V = m*R*T
(p=pressure, V=volume, m=mass, R=specific gas constant, T=temperature)
m = p*V / R*T

V = 7.70×10^-2m^3
R = RO2 = 259.8 J/kg*K

A) the initial mass 'm1'
p1 = 3.05×10^5Pa
T1 = 39 C = 39 + 273.1K = 312.1 K

m1 = p1 * V / R * T1 = 0.2896 kg = 289.6 g
(units: Pa = N/m² , J = N * m)

B) the mass of oxygen that has leakes out 'm12'
m12 = m1 - m2
p2 = 1.9×10^5Pa
T2 = 20.8C = 293.9 K

m2 = p2 * V / R * T2 = 0.1916 kg

m12 = m1 - m2 = 0.098kg = 98 g

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