J.R. S. answered • 03/05/18

Ph.D. in Biochemistry--University Professor--Chemistry Tutor

Nanda J.

03/05/18

Nanda J.

03/05/18

J.R. S.

03/05/18

Nanda J.

03/05/18

J.R. S.

03/05/18

Nanda J.

asked • 03/05/18At 27 degree centigrade a cylinder of 20 litres capacity contains three gases HE O2 N2 their masses are 0.502g 0.250g 1g respectively if all these gases behave ideally calculate partial pressures of each gas as well as total pressure..

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J.R. S. answered • 03/05/18

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Ph.D. in Biochemistry--University Professor--Chemistry Tutor

Moles He = 0.502 g x 1 mol/4 g = 0.1255 moles

Moles O2 = 0.250 g x1 mol/32 g = 0.00681 moles

Moles N2 = 1 g x 1 mol/28g = 0.0357 moles

total moles = 0.168 moles

Total pressure PV = nRT and P = nRT/V

P = (0.168)(0.0821)(300)/20 = 0.207 atm = TOTAL PRESSURE

Partial pressure is determined from mole fraction of each gas

mole fraction He = 0.1255/0.168 = 0.747

partial pressure He = 0.747 x 0.207 atm = 0.155 atm

mole fraction O2 = 0.0068/0.168 = 0.0404

partial pressure O2 = 0.0404 x 0.207 atm = 0.0084 atm

mole fraction N2 = 0.0357/0.168 = 0.212

partial pressure N2 = 0.212 x 0.207 atm = 0.044 atm

Nanda J.

Sir thanks for the reply but in second line

Moles O2 = 0.250 g x1 mol/32 g = 0.0078125 moles?? Correction Sir?

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03/05/18

Nanda J.

Moles O2 = 0.250 g x1 mol/32 g = 0.00781 moles : -Correction*

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03/05/18

J.R. S.

tutor

Sorry. What’s the problem/question?

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03/05/18

Nanda J.

Sir in second line of your answer you have written

Moles O2 = 0.250 g x1 mol/32 g = 0.00681 moles

But sir 0.250/32 is 0.00781 and not 0.00681..

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03/05/18

J.R. S.

tutor

You are correct. A typo error on my part.

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03/05/18

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Nanda J.

03/05/18