Chapter 10, Problem 57GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Complete the following table:

Interpretation Introduction

Interpretation: For the given table that contains set of gases with their pressure values in given units should be converted in the other given units.

Concept introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas.  Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained.  It is referred as ideal gas equation.

nTPV = RnTPPV = nRTwhere,n = molesofgasP = pressureT = temperatureR = gas constant

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties.  At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to its volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

Explanation

At standard conditions there pressure is expressed as 1 atm which is also expressed in terms of â€ŠmmÂ Hgâ€Šâ€Š, kPaâ€Š and â€Šbar.

â€‚Â 1atmÂ =Â 760Â mmÂ Hg1atmÂ =Â 101,325Â Paâ€Šâ€Š=â€Šâ€Š101.325Â kPaâ€Š1atmÂ =Â 1.013Â bar

It is given that partial pressure of N2 is 593â€Šâ€ŠmmÂ Hgâ€Šâ€Š which is expressed in terms of atm,kPaâ€Š and â€Šbar as follows by using the above units conversion table.

â€‚Â N2=â€Šâ€Š593Â â€ŠmmÂ Hgâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=â€Šâ€Š593Â â€ŠmmÂ Hg760â€Šâ€Šâ€ŠmmÂ HgÃ—1atmâ€Šâ€Š=0.780â€Šâ€ŠatmN2=0.780â€Šâ€Šatmâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=0.780â€Šâ€ŠatmÃ—101.325kPaâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=79.03kPaN2=0.780â€Šâ€ŠatmÃ—1.013â€Šâ€Šbarâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=0.790Â bar

â€‚Â H2=â€Šâ€Š133Â â€Šbarâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=â€Šâ€Š133Â â€Šbar1.013Â barÃ—1atmâ€Šâ€Š=131â€Šâ€ŠatmH2=Â 131â€Šâ€Šatmâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=Â 131â€Šâ€ŠatmÃ—101.325kPaâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=Â 13273

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