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Normal boiling point of water is 373K (at 760mm). Vapor pressure of water at 298K is 23mm. If enthalpy of evaporation is 40656 J/mol, boiling point of water 23mm pressure will be:
[R = 8.314821 JK-1mol-1]
[R = 8.314821 JK-1mol-1]
- a)250 K
- b)294 K
- c)51.6 K
- d)12.5 K
Correct answer is option 'B'. Can you explain this answer?
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Normal boiling point of water is 373K (at 760mm). Vapor pressure of wa...
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Normal boiling point of water is 373K (at 760mm). Vapor pressure of wa...
Given Information:
- Normal boiling point of water (at 760 mmHg) = 373 K
- Vapor pressure of water at 298 K = 23 mmHg
- Enthalpy of evaporation = 40656 J/mol
- Gas constant (R) = 8.314821 J K-1 mol-1
Calculating the boiling point of water at 23 mmHg pressure:
- Use the Clausius-Clapeyron equation: ln(P2/P1) = (ΔH/R) x (1/T1 - 1/T2)
- P1 = 760 mmHg, T1 = 373 K, P2 = 23 mmHg, ΔH = 40656 J/mol, R = 8.314821 J K-1 mol-1
- Convert pressures to atm: P1 = 1 atm, P2 = 23/760 atm
- Convert ΔH to J: ΔH = 40656 J/mol
- Solve for T2 (boiling point at 23 mmHg pressure) using the equation
- T2 ≈ 294 K
Therefore, the boiling point of water at 23 mmHg pressure will be approximately 294 K. Hence, the correct answer is option B.
- Normal boiling point of water (at 760 mmHg) = 373 K
- Vapor pressure of water at 298 K = 23 mmHg
- Enthalpy of evaporation = 40656 J/mol
- Gas constant (R) = 8.314821 J K-1 mol-1
Calculating the boiling point of water at 23 mmHg pressure:
- Use the Clausius-Clapeyron equation: ln(P2/P1) = (ΔH/R) x (1/T1 - 1/T2)
- P1 = 760 mmHg, T1 = 373 K, P2 = 23 mmHg, ΔH = 40656 J/mol, R = 8.314821 J K-1 mol-1
- Convert pressures to atm: P1 = 1 atm, P2 = 23/760 atm
- Convert ΔH to J: ΔH = 40656 J/mol
- Solve for T2 (boiling point at 23 mmHg pressure) using the equation
- T2 ≈ 294 K
Therefore, the boiling point of water at 23 mmHg pressure will be approximately 294 K. Hence, the correct answer is option B.
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Normal boiling point of water is 373K (at 760mm). Vapor pressure of water at 298K is 23mm. If enthalpy of evaporation is 40656 J/mol, boiling point of water 23mm pressure will be:[R = 8.314821 JK-1mol-1]a)250 Kb)294 Kc)51.6 Kd)12.5 KCorrect answer is option 'B'. Can you explain this answer?
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Normal boiling point of water is 373K (at 760mm). Vapor pressure of water at 298K is 23mm. If enthalpy of evaporation is 40656 J/mol, boiling point of water 23mm pressure will be:[R = 8.314821 JK-1mol-1]a)250 Kb)294 Kc)51.6 Kd)12.5 KCorrect answer is option 'B'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Normal boiling point of water is 373K (at 760mm). Vapor pressure of water at 298K is 23mm. If enthalpy of evaporation is 40656 J/mol, boiling point of water 23mm pressure will be:[R = 8.314821 JK-1mol-1]a)250 Kb)294 Kc)51.6 Kd)12.5 KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Normal boiling point of water is 373K (at 760mm). Vapor pressure of water at 298K is 23mm. If enthalpy of evaporation is 40656 J/mol, boiling point of water 23mm pressure will be:[R = 8.314821 JK-1mol-1]a)250 Kb)294 Kc)51.6 Kd)12.5 KCorrect answer is option 'B'. Can you explain this answer?.
Normal boiling point of water is 373K (at 760mm). Vapor pressure of water at 298K is 23mm. If enthalpy of evaporation is 40656 J/mol, boiling point of water 23mm pressure will be:[R = 8.314821 JK-1mol-1]a)250 Kb)294 Kc)51.6 Kd)12.5 KCorrect answer is option 'B'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Normal boiling point of water is 373K (at 760mm). Vapor pressure of water at 298K is 23mm. If enthalpy of evaporation is 40656 J/mol, boiling point of water 23mm pressure will be:[R = 8.314821 JK-1mol-1]a)250 Kb)294 Kc)51.6 Kd)12.5 KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Normal boiling point of water is 373K (at 760mm). Vapor pressure of water at 298K is 23mm. If enthalpy of evaporation is 40656 J/mol, boiling point of water 23mm pressure will be:[R = 8.314821 JK-1mol-1]a)250 Kb)294 Kc)51.6 Kd)12.5 KCorrect answer is option 'B'. Can you explain this answer?.
Solutions for Normal boiling point of water is 373K (at 760mm). Vapor pressure of water at 298K is 23mm. If enthalpy of evaporation is 40656 J/mol, boiling point of water 23mm pressure will be:[R = 8.314821 JK-1mol-1]a)250 Kb)294 Kc)51.6 Kd)12.5 KCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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