Chapter 10, Problem 77GQ

### 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

# Chlorine dioxide, ClO2, reacts with fluorine to give a new gas that contains Cl, O, and F. In an experiment, you find that 0.150 g of this new gas has a pressure of 17.2 mm Hg in a 1850-mL flask at 21 °C. What is the identity of the unknown gas?

Interpretation Introduction

Interpretation: The identity of the gas present in the given chemical reaction under given conditions should be determined.

Concept introduction:

Ideal gas Equation:

Any gas can be 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 = moles of gasP = 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 the 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.

Van der Waal’s gas equation:

The van der Waal equation describes the ideal gas as it approaches to zero.  The van der Waal equation contains correction terms a and b for the intermolecular forces and molecular size respectively.

The van der Waal equation is as follows,

[P+a(nV)2](Vnb)=RT

Explanation

â€‚Â PVÂ =Â nRTâ€‰â€‰â€‰â€‰nâ€‰=â€Šâ€ŠmassmolarÂ massPVÂ =Â massmolarÂ massRTMolarÂ massÂ =wRTPVâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€Šâ€Š=â€Šâ€Š0.150gÃ—0.0821Ã—294.1517.2760Ã—18

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