The normal boiling point of CCl 4 should be predicted. Concept Introduction: The Clausius Clapeyron equation is given as follows: ln P 2 P 1 = − Δ H v a p R ( 1 T 2 − 1 T 1 ) Where, P 2 is the vapour pressure at T 2 P 1 is the vapour pressure at T 1 Δ H v a p is the enthalpy of vaporization R is the Gas constant.

Question

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Answer 1

Question

Chapter 16, Problem 87E

Interpretation Introduction

Interpretation:

The normal boiling point of CCl4 should be predicted.

Concept Introduction:

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Where,

P2 is the vapour pressure at T2

P1 is the vapour pressure at T1

ΔHvap is the enthalpy of vaporization

R is the Gas constant.

Expert Solution & Answer

Answer to Problem 87E

The normal boiling point of CCl4 is 76.9°C.

Explanation of Solution

Given information:

P2 is 836 torr at T2=80.°C=80+273=353 K
P1 is 213 torr at T1=40.°C=40+273=313 K

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Substitute the given values in above formula.

ln( 836 torr 213 torr)=−ΔH vap8.314 J K −1 mol −1(1 353 K−1 313 K)1.3673=−ΔH vap8.314 J K −1 mol −1(−3.62× 10 −4 K −1)ΔHvap=1.3673( 8.314 J K −1 mol −1 )3.62× 10 −4 K −1=31402.5J mol−1

Therefore, the value of ΔHvap is 31402.5J mol−1.

Substitute T1=313 K , P1=213 torr , ΔHvap=31402.5J mol−1 and P2=1 atm or 760 torr in Clausius Clapeyron equation.

ln( 760 torr 213 torr)=−31402.5 J mol −18.314 J K −1 mol −1(1 T 2 −1 313 K)1.272=(−3777.06 K)(1 T 2 −1 313 K)1T2−1313 K=−1.2723777.06 K1T2=1313 K−1.2723777.06 K1T2=0.002858 K−1T2=10.002858 K −1=349.9 K=76.9°C

Therefore, the normal boiling point of CCl4 is 76.9°C.

Conclusion

The normal boiling point of CCl4 is 76.9°C.

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Answer 2

Question

Chapter 16, Problem 87E

Interpretation Introduction

Interpretation:

The normal boiling point of CCl4 should be predicted.

Concept Introduction:

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Where,

P2 is the vapour pressure at T2

P1 is the vapour pressure at T1

ΔHvap is the enthalpy of vaporization

R is the Gas constant.

Expert Solution & Answer

Answer to Problem 87E

The normal boiling point of CCl4 is 76.9°C.

Explanation of Solution

Given information:

P2 is 836 torr at T2=80.°C=80+273=353 K
P1 is 213 torr at T1=40.°C=40+273=313 K

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Substitute the given values in above formula.

ln( 836 torr 213 torr)=−ΔH vap8.314 J K −1 mol −1(1 353 K−1 313 K)1.3673=−ΔH vap8.314 J K −1 mol −1(−3.62× 10 −4 K −1)ΔHvap=1.3673( 8.314 J K −1 mol −1 )3.62× 10 −4 K −1=31402.5J mol−1

Therefore, the value of ΔHvap is 31402.5J mol−1.

Substitute T1=313 K , P1=213 torr , ΔHvap=31402.5J mol−1 and P2=1 atm or 760 torr in Clausius Clapeyron equation.

ln( 760 torr 213 torr)=−31402.5 J mol −18.314 J K −1 mol −1(1 T 2 −1 313 K)1.272=(−3777.06 K)(1 T 2 −1 313 K)1T2−1313 K=−1.2723777.06 K1T2=1313 K−1.2723777.06 K1T2=0.002858 K−1T2=10.002858 K −1=349.9 K=76.9°C

Therefore, the normal boiling point of CCl4 is 76.9°C.

Conclusion

The normal boiling point of CCl4 is 76.9°C.

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Answer 3

Question

Chapter 16, Problem 87E

Interpretation Introduction

Interpretation:

The normal boiling point of CCl4 should be predicted.

Concept Introduction:

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Where,

P2 is the vapour pressure at T2

P1 is the vapour pressure at T1

ΔHvap is the enthalpy of vaporization

R is the Gas constant.

Expert Solution & Answer

Answer to Problem 87E

The normal boiling point of CCl4 is 76.9°C.

Explanation of Solution

Given information:

P2 is 836 torr at T2=80.°C=80+273=353 K
P1 is 213 torr at T1=40.°C=40+273=313 K

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Substitute the given values in above formula.

ln( 836 torr 213 torr)=−ΔH vap8.314 J K −1 mol −1(1 353 K−1 313 K)1.3673=−ΔH vap8.314 J K −1 mol −1(−3.62× 10 −4 K −1)ΔHvap=1.3673( 8.314 J K −1 mol −1 )3.62× 10 −4 K −1=31402.5J mol−1

Therefore, the value of ΔHvap is 31402.5J mol−1.

Substitute T1=313 K , P1=213 torr , ΔHvap=31402.5J mol−1 and P2=1 atm or 760 torr in Clausius Clapeyron equation.

ln( 760 torr 213 torr)=−31402.5 J mol −18.314 J K −1 mol −1(1 T 2 −1 313 K)1.272=(−3777.06 K)(1 T 2 −1 313 K)1T2−1313 K=−1.2723777.06 K1T2=1313 K−1.2723777.06 K1T2=0.002858 K−1T2=10.002858 K −1=349.9 K=76.9°C

Therefore, the normal boiling point of CCl4 is 76.9°C.

Conclusion

The normal boiling point of CCl4 is 76.9°C.

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Answer 4

Question

Chapter 16, Problem 87E

Interpretation Introduction

Interpretation:

The normal boiling point of CCl4 should be predicted.

Concept Introduction:

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Where,

P2 is the vapour pressure at T2

P1 is the vapour pressure at T1

ΔHvap is the enthalpy of vaporization

R is the Gas constant.

Expert Solution & Answer

Answer to Problem 87E

The normal boiling point of CCl4 is 76.9°C.

Explanation of Solution

Given information:

P2 is 836 torr at T2=80.°C=80+273=353 K
P1 is 213 torr at T1=40.°C=40+273=313 K

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Substitute the given values in above formula.

ln( 836 torr 213 torr)=−ΔH vap8.314 J K −1 mol −1(1 353 K−1 313 K)1.3673=−ΔH vap8.314 J K −1 mol −1(−3.62× 10 −4 K −1)ΔHvap=1.3673( 8.314 J K −1 mol −1 )3.62× 10 −4 K −1=31402.5J mol−1

Therefore, the value of ΔHvap is 31402.5J mol−1.

Substitute T1=313 K , P1=213 torr , ΔHvap=31402.5J mol−1 and P2=1 atm or 760 torr in Clausius Clapeyron equation.

ln( 760 torr 213 torr)=−31402.5 J mol −18.314 J K −1 mol −1(1 T 2 −1 313 K)1.272=(−3777.06 K)(1 T 2 −1 313 K)1T2−1313 K=−1.2723777.06 K1T2=1313 K−1.2723777.06 K1T2=0.002858 K−1T2=10.002858 K −1=349.9 K=76.9°C

Therefore, the normal boiling point of CCl4 is 76.9°C.

Conclusion

The normal boiling point of CCl4 is 76.9°C.

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Answer 5

Chapter 16, Problem 87E

Interpretation Introduction

Interpretation:

The normal boiling point of CCl4 should be predicted.

Concept Introduction:

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Where,

P2 is the vapour pressure at T2

P1 is the vapour pressure at T1

ΔHvap is the enthalpy of vaporization

R is the Gas constant.

Expert Solution & Answer

Answer to Problem 87E

The normal boiling point of CCl4 is 76.9°C.

Explanation of Solution

Given information:

P2 is 836 torr at T2=80.°C=80+273=353 K
P1 is 213 torr at T1=40.°C=40+273=313 K

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Substitute the given values in above formula.

ln( 836 torr 213 torr)=−ΔH vap8.314 J K −1 mol −1(1 353 K−1 313 K)1.3673=−ΔH vap8.314 J K −1 mol −1(−3.62× 10 −4 K −1)ΔHvap=1.3673( 8.314 J K −1 mol −1 )3.62× 10 −4 K −1=31402.5J mol−1

Therefore, the value of ΔHvap is 31402.5J mol−1.

Substitute T1=313 K , P1=213 torr , ΔHvap=31402.5J mol−1 and P2=1 atm or 760 torr in Clausius Clapeyron equation.

ln( 760 torr 213 torr)=−31402.5 J mol −18.314 J K −1 mol −1(1 T 2 −1 313 K)1.272=(−3777.06 K)(1 T 2 −1 313 K)1T2−1313 K=−1.2723777.06 K1T2=1313 K−1.2723777.06 K1T2=0.002858 K−1T2=10.002858 K −1=349.9 K=76.9°C

Therefore, the normal boiling point of CCl4 is 76.9°C.

Conclusion

The normal boiling point of CCl4 is 76.9°C.

Answer 6

Chapter 16, Problem 87E

Interpretation Introduction

Interpretation:

The normal boiling point of CCl4 should be predicted.

Concept Introduction:

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Where,

P2 is the vapour pressure at T2

P1 is the vapour pressure at T1

ΔHvap is the enthalpy of vaporization

R is the Gas constant.

Expert Solution & Answer

Answer to Problem 87E

The normal boiling point of CCl4 is 76.9°C.

Explanation of Solution

Given information:

P2 is 836 torr at T2=80.°C=80+273=353 K
P1 is 213 torr at T1=40.°C=40+273=313 K

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Substitute the given values in above formula.

ln( 836 torr 213 torr)=−ΔH vap8.314 J K −1 mol −1(1 353 K−1 313 K)1.3673=−ΔH vap8.314 J K −1 mol −1(−3.62× 10 −4 K −1)ΔHvap=1.3673( 8.314 J K −1 mol −1 )3.62× 10 −4 K −1=31402.5J mol−1

Therefore, the value of ΔHvap is 31402.5J mol−1.

Substitute T1=313 K , P1=213 torr , ΔHvap=31402.5J mol−1 and P2=1 atm or 760 torr in Clausius Clapeyron equation.

ln( 760 torr 213 torr)=−31402.5 J mol −18.314 J K −1 mol −1(1 T 2 −1 313 K)1.272=(−3777.06 K)(1 T 2 −1 313 K)1T2−1313 K=−1.2723777.06 K1T2=1313 K−1.2723777.06 K1T2=0.002858 K−1T2=10.002858 K −1=349.9 K=76.9°C

Therefore, the normal boiling point of CCl4 is 76.9°C.

Conclusion

The normal boiling point of CCl4 is 76.9°C.

Answer 7

Chapter 16, Problem 87E

Interpretation Introduction

Interpretation:

The normal boiling point of CCl4 should be predicted.

Concept Introduction:

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Where,

P2 is the vapour pressure at T2

P1 is the vapour pressure at T1

ΔHvap is the enthalpy of vaporization

R is the Gas constant.

Answer 8

Expert Solution & Answer

Answer to Problem 87E

The normal boiling point of CCl4 is 76.9°C.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

P2 is 836 torr at T2=80.°C=80+273=353 K
P1 is 213 torr at T1=40.°C=40+273=313 K

The Clausius Clapeyron equation is given as follows:

lnP2P1=−ΔHvapR(1T2−1T1)

Substitute the given values in above formula.

ln( 836 torr 213 torr)=−ΔH vap8.314 J K −1 mol −1(1 353 K−1 313 K)1.3673=−ΔH vap8.314 J K −1 mol −1(−3.62× 10 −4 K −1)ΔHvap=1.3673( 8.314 J K −1 mol −1 )3.62× 10 −4 K −1=31402.5J mol−1

Therefore, the value of ΔHvap is 31402.5J mol−1.

Substitute T1=313 K , P1=213 torr , ΔHvap=31402.5J mol−1 and P2=1 atm or 760 torr in Clausius Clapeyron equation.

ln( 760 torr 213 torr)=−31402.5 J mol −18.314 J K −1 mol −1(1 T 2 −1 313 K)1.272=(−3777.06 K)(1 T 2 −1 313 K)1T2−1313 K=−1.2723777.06 K1T2=1313 K−1.2723777.06 K1T2=0.002858 K−1T2=10.002858 K −1=349.9 K=76.9°C