An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32μ).
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Explanation:
Volume of oxygen, V1 = 30 litres = 30 × 10–3 m3
Gauge pressure, P1 = 15 atm = 15 × 1.013 × 105 Pa
Temperature, T1 = 27°C = 300 K
Universal gas constant, R = 8.314 J mole–1 K–1
Let the initial number of moles of oxygen gas in the cylinder be n1.
The gas equation is given as:
P1V1 = n1RT1
∴ n1 = P1V1/ RT1
= (15.195 × 105 × 30 × 10-3) / (8.314 × 300) = 18.276
But n1 = m1 / M
Where,
m1 = Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
∴m1 = n1M = 18.276 × 32 = 584.84 g
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume, V2 = 30 litres = 30 × 10–3 m3
Gauge pressure, P2 = 11 atm = 11 × 1.013 × 105 Pa
Temperature, T2 = 17°C = 290 K
Let n2 be the number of moles of oxygen left in the cylinder.
The gas equation is given as:
P2V2 = n2RT2
∴ n2 = P2V2/ RT2
= (11.143 × 105 × 30 × 10-3) / (8.314 × 290) = 13.86
But n2 = m2 / M
Where,
m2 is the mass of oxygen remaining in the cylinder
∴ m2 = n2M = 13.86 × 32 = 453.1 g
The mass of oxygen taken out of the cylinder is given by the relation:
Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder
= m1 – m2
= 584.84 g – 453.1 g
= 131.74 g
= 0.131 kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.
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Volume of oxygen, V1= 30 litres =30×10−3m3
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105Pa
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300K
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:P1V1=n1RT1
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:P1V1=n1RT1∴n1=P1V1/RT1
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:P1V1=n1RT1∴n1=P1V1/RT1=(15.195×105×30×10−3)/(8.314×300)=18.276
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:P1V1=n1RT1∴n1=P1V1/RT1=(15.195×105×30×10−3)/(8.314×300)=18.276But n1=m1/M
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:P1V1=n1RT1∴n1=P1V1/RT1=(15.195×105×30×10−3)/(8.314×300)=18.276But n1=m1/MWhere,
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:P1V1=n1RT1∴n1=P1V1/RT1=(15.195×105×30×10−3)/(8.314×300)=18.276But n1=m1/MWhere, m1= Initial mass of oxygen
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:P1V1=n1RT1∴n1=P1V1/RT1=(15.195×105×30×10−3)/(8.314×300)=18.276But n1=m1/MWhere, m1= Initial mass of oxygen M = Molecular mass of oxygen = 32 g
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:P1V1=n1RT1∴n1=P1V1/RT1=(15.195×105×30×10−3)/(8.314×300)=18.276But n1=m1/MWhere, m1= Initial mass of oxygen M = Molecular mass of oxygen = 32 g∴m1=N1M=18.276×32=584.84g
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:P1V1=n1RT1∴n1=P1V1/RT1=(15.195×105×30×10−3)/(8.314×300)=18.276But n1=m1/MWhere, m1= Initial mass of oxygen M = Molecular mass of oxygen = 32 g∴m1=N1M=18.276×32=584.84gAfter some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume of oxygen, V1= 30 litres =30×10−3m3 Gauge pressure, P1= 15 atm =15×1.013×105PaTemperature, T1=27oC=300KUniversal gas constant, R=8.314Jmol−1K−1Let the initial number of moles of oxygen gas in the cylinder be n1 .The gas equation is given as:P1V1=n1RT1∴n1=P1V1/RT1=(15.195×105×30×10−3)/(8.314×300)=18.276But n1=m1/MWhere, m1= Initial mass of oxygen M = Molecular mass of oxygen = 32 g∴m1=N1M=18.276×32=584.84gAfter some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.Volume, V2= 30 litres =
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