Chapter 10, Problem 54PS

### 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

# You want to store 165 g of CO2 gas in a 12.5-L tank at room temperature (25 °C). Calculate the pressure the gas would have using (a) the ideal gas law and (b) the van der Waals equation. (For CO2, a = 3.59 atm • L2/mol2 and b = 0.0427 L/mol.)

(a)

Interpretation Introduction

Interpretation: For the gas under given conditions the pressure for the gas has to be determined by using ideal gas equation.

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 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.

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

Given,

â€‚Â Volume,Â Vâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€ŠÂ =Â 12.5Â LTemperature,Â Tâ€Šâ€Šâ€Š=â€Šâ€Š25oCâ€Šâ€Šâ€Š=â€Šâ€Šâ€Š298.15Kâ€Šâ€Šâ€Šâ€ŠmassÂ â€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=Â 165Â gGasÂ constant,Â RÂ =Â 0.0821Â LÂ atmÂ mol-1K-1Pressure,Â Pâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€ŠÂ =Â ?

First convert the given amount of gas into moles which is achieved as follow,

â€‚Â molesÂ =Â massmolarÂ massâ€Šâ€Š=â€Šâ€Š165g44

(b)

Interpretation Introduction

Interpretation: The pressure for the gas has be determined by the van der Waals equation.

Concept introduction:

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.

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

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