The barometric pressure at the sea level is 760mmHg. The pressure of the pressure cooker is 1.75 atm. The normal boiling point temperature difference between the boiling water in an open pot and in pressure cooker is to be determined. The heat of vaporization of water is ΔH vap =40.7kJ/mol. Concept introduction: The Clausius − Clapeyron equation is as follows: ln P 1 P 2 = − Δ H v a p R ( 1 T 1 − 1 T 2 ) Here, P 1 and P 2 = vapor pressure T 1 and T 2 = Temperature R = gas constant ΔH vap = heat of vaporization The boiling point temperature difference is - Δ T b = T 2 − T 1

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MISSED THIS? Read Section 19.9 (Pages 878-881); Watch IWE 19.10 Consider the following reaction: CH3OH(g) CO(g) + 2H2(g) (Note that AG,CH3OH(g) = -162.3 kJ/mol and AG,co(g)=-137.2 kJ/mol.) Part A Calculate AG for this reaction at 25 °C under the following conditions: PCH₂OH Pco PH2 0.815 atm = 0.140 atm 0.170 atm Express your answer in kilojoules to three significant figures. Ο ΑΣΦ AG = -150 Submit Previous Answers Request Answer □? kJ × Incorrect; Try Again; 2 attempts remaining Calculate the free energy change under nonstandard conditions (AGrxn) by using the following relationship: AGrxn = AGrxn + RTInQ, AGxn+RTInQ, where AGxn is the standard free energy change, R is the ideal gas constant, T is the temperature in kelvins, a is the reaction quotient. Provide Feedback Next >

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Chapter 9, Problem 12QAP

Interpretation Introduction

Interpretation:

The barometric pressure at the sea level is 760mmHg. The pressure of the pressure cooker is 1.75 atm. The normal boiling point temperature difference between the boiling water in an open pot and in pressure cooker is to be determined. The heat of vaporization of water is ΔH_vap =40.7kJ/mol.

Concept introduction:

The Clausius − Clapeyron equation is as follows:

lnP1P2=−ΔHvapR(1T1−1T2)

Here,

P₁ and P₂ = vapor pressure

T₁ and T₂ = Temperature

R = gas constant

ΔH_vap = heat of vaporization

The boiling point temperature difference is -

ΔTb=T2−T1

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