CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
14th Edition
ISBN: 9781259137815
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 23, Problem 81AP
Interpretation Introduction

Interpretation:

The pressure is to be calculated.

Concept Introduction:

The pressure is an expression of force exerted on a surface per unit area.

The relation between pressure and density of a gas is as

P=dRTM.

Here, P is the pressure, d is the density, R is the universal gas constant, M is molar mass, and T is the temperature. The conversion of temperature from degree Celsius to Kelvin can be done by using the formula: T(K)=T(°C)+273.

The density of O2 in KO2 using the mass percentage of O2 in the compound as:

massofO2massofKO2×density.

The relationship between liters and cubic meters can be expressed as:

1L=1000cm3.

To convert cubic meters to liters, conversion factor is 1L1000cm3.

Expert Solution & Answer
Check Mark

Answer to Problem 81AP

Solution: 727 atm

Explanation of Solution

Given information:

Density, d=2.15 g/cm3

Temperature, T=20oC

The value of the gas constant is 0.0821 Latm/molK.

The molar mass of oxygen is 32.00 g/mol.

The molar mass of KO2 is 71.10 g/mol.

The temperature is 80.1°C.

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

T(K)=T(°C)+273=(20+273)=293 K.

Calculate the density of O2 in KO2 using the mass percentage of O2 in the compound as:

massofO2massofKO2×density.

32.00gO271.10gKO2×2.15g KO21cm3=0.968g O2 /cm3.

In one liter, there are thousand cubic centimeters.

Convert cubic centimeter to liter as:

0.968/cm3=(0.9681cm3)(1000cm31L)=968/L.

Calculate the pressure from the relationship between pressure and molar mass as

P=dRTM.

Substitute 0.0821 Latm/molK for R, 968/L for d, 293 K for T, and 32.00 g/mol for M in the above equation.

P=968/L(0.0821 Latm/molK)(293K)32.00 g/mol=727atm.

Conclusion

The calculated pressureis 727atm.

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Chapter 23 Solutions

CHEMISTRY >CUSTOM<

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