Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
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Chapter 1, Problem 1C.6E

(a)

n der Waal  ues in the perfect gas equation as follows,

to be detetrmined.

Interpretation Introduction

Interpretation:

The pressure exerted by 1.0 mol of C2H6 behaving as a perfect gas under the given conditions has to be determined.

Concept introduction:

Ideal gas equation,

PV=nRTWhere,P:PressureV:Volumen:NumberofmolesR:GasconstantT:Temperature

Van der Waals equation:

The deviation from ideal gas law was accounted using two constants, a and b and was given by an equation called van der Waals equation which is similar to ideal gas law as follows,

   (P+n2aV2)(v-nb)=nRT

  Where,

a and b are constants

P is the pressure

V is the volume

T is the temperature

R is molar gas constant

n is the mole

Using van der Waal’s equation, the pressure can be determined as,

P=nRTVnba(nV)2

(a)

n der Waal  ues in the perfect gas equation as follows,

to be detetrmined.

Expert Solution
Check Mark

Explanation of Solution

The pressure exerted by 1.0 mol of C2H6 behaving as a perfect gas under the given conditions can determined as follows,

  1. (i) 273.15 K in 22.414dm3

Conversions:

1dm=10cm,1dm3=1000cm322.414dm3=22.414L

n=1molR=0.0821LatmK-1mol-1T=273.15KV=22.414L

Substituting the known values in the perfect gas equation as follows,

P=nRTV=(1mol)×(0.0821)×(273.15K)22.414L=1atm

  1. (ii) At 1000 K in 100cm3

100cm3=100mL=0.1L

n=1molR=0.0821LatmK-1mol-1T=1000KV=0.1L

Substituting the known values in the perfect gas equation as follows,

P=nRTV=(1mol)×(0.0821)×(1000K)0.1L=820atm

(b)

n der Waal  ues in the perfect gas equation as follows,

to be detetrmined.

Interpretation Introduction

Interpretation:

The pressure exerted by 1.0 mol of C2H6 behaving as a van der Waal gas under the given conditions has to be determined.

Concept introduction:

Ideal gas equation,

PV=nRTWhere,P:PressureV:Volumen:NumberofmolesR:GasconstantT:Temperature

Van der Waals equation:

The deviation from ideal gas law was accounted using two constants, a and b and was given by an equation called van der Waals equation which is similar to ideal gas law as follows,

   (P+n2aV2)(v-nb)=nRT

  Where,

a and b are constants

P is the pressure

V is the volume

T is the temperature

R is molar gas constant

n is the mole

Using van der Waal’s equation, the pressure can be determined as,

P=nRTVnba(nV)2

(b)

n der Waal  ues in the perfect gas equation as follows,

to be detetrmined.

Expert Solution
Check Mark

Explanation of Solution

The pressure exerted by 1.0 mol of C2H6 behaving as a van der Waal gas under the given conditions can be determined as follows,

  1. (i) 273.15 K in 22.414dm3

Conversions:

1dm=10cm,1dm3=1000cm322.414dm3=22.414L

n=1molR=0.0821LatmK-1mol-1T=273.15KV=22.414L

Using van der Waal’s equation, the pressure can be determined as,

P=nRTVnba(nV)2

a=5.507atmdm6mol1b=6.51×102dm3mol1=6.51×102Lmol1

Substituting the known values in the van der Waal gas equation as follows,

P=nRTVnbn2aV2=(1mol)(0.0821)(273.15)22.414(1mol×6.51×102)(1)2×5.507atmdm6mol1(22.414dm3)2=0.99atm

  1. (ii) At 1000 K in 100cm3

100cm3=100mL=0.1L

n=1molR=0.0821LatmK-1mol-1T=1000KV=0.1L

a=5.507atmdm6mol1b=6.51×102dm3mol1=6.51×102Lmol1

Substituting the known values in the van der Waal gas equation as follows,

P=nRTVnbn2aV2=(1mol)(0.0821)(1000K)0.1L(1mol×6.51×102)(1)2×5.507atmdm6mol1(0.1dm3)2=1.8×103atm

n der Waal  ues in the perfect gas equation as follows,

to be detetrmined.

n der Waal  ues in the perfect gas equation as follows,

to be detetrmined.

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