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The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23 mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:
- a)250 K
- b)294 K
- c)230 K
- d)290 K
Correct answer is option 'B'. Can you explain this answer?
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The normal boiling point of water is 373 K. Vapour pressure of water a...
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The normal boiling point of water is 373 K. Vapour pressure of water a...
To solve this problem, we need to use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and the enthalpy of vaporization.
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
where P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively, ΔHvap is the enthalpy of vaporization, R is the gas constant, and T1 and T2 are the corresponding temperatures.
In this problem, we are given the normal boiling point of water (373 K), the vapor pressure of water (23 mm Hg), and the enthalpy of vaporization of water (40.67 kJ/mol). We are asked to find the temperature (T) at which the vapor pressure of water is 23 mm Hg.
Let's solve the equation:
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (1/T - 1/373)
Simplifying the equation, we get:
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (1/T - 1/373)
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373/T - 1/373)
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373 - T/373)
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373 - T/373)
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373 - T/373)
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373 - T/373)
Now, let's solve for T:
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373 - T/373)
Using a calculator or software, we find that T ≈ 294 K.
Therefore, the correct answer is option B) 294 K.
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
where P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively, ΔHvap is the enthalpy of vaporization, R is the gas constant, and T1 and T2 are the corresponding temperatures.
In this problem, we are given the normal boiling point of water (373 K), the vapor pressure of water (23 mm Hg), and the enthalpy of vaporization of water (40.67 kJ/mol). We are asked to find the temperature (T) at which the vapor pressure of water is 23 mm Hg.
Let's solve the equation:
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (1/T - 1/373)
Simplifying the equation, we get:
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (1/T - 1/373)
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373/T - 1/373)
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373 - T/373)
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373 - T/373)
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373 - T/373)
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373 - T/373)
Now, let's solve for T:
ln(23/760) = (-40.67 * 10^3 J/mol / (8.314 J/(mol*K))) * (373 - T/373)
Using a calculator or software, we find that T ≈ 294 K.
Therefore, the correct answer is option B) 294 K.
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The normal boiling point of water is 373 K. Vapour pressure of water a...
373 Kelvin and about here the problem of 5opical reference and region and the and is b
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The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:a)250 Kb)294 Kc)230 Kd)290 KCorrect answer is option 'B'. Can you explain this answer?
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The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:a)250 Kb)294 Kc)230 Kd)290 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 The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:a)250 Kb)294 Kc)230 Kd)290 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 The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:a)250 Kb)294 Kc)230 Kd)290 KCorrect answer is option 'B'. Can you explain this answer?.
The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:a)250 Kb)294 Kc)230 Kd)290 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 The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:a)250 Kb)294 Kc)230 Kd)290 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 The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:a)250 Kb)294 Kc)230 Kd)290 KCorrect answer is option 'B'. Can you explain this answer?.
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The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:a)250 Kb)294 Kc)230 Kd)290 KCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:a)250 Kb)294 Kc)230 Kd)290 KCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of The normal boiling point of water is 373 K. Vapour pressure of water at temperature T is 23mm Hg. If the enthalpy of vaporization is 40.67 kJ/mol, then temperature T would be:a)250 Kb)294 Kc)230 Kd)290 KCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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