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Passage II
Variation of log10 p with 1/T is given in the following graph (pressure in bar)
 
Q. Enthalpy of vaporisation based on this data is
  • a)
    460.6 cal mol -1
  • b)
    -460.6 cal mol-1
  • c)
    200.0 cal mol -1
  • d)
    - 200.0 cal mol -1
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Passage IIVariation of log10p with1/Tis given in the following graph (...
We have, log p2/p1 = ∆Hvap / 2.303 R [1/T2-1/T1]  
Or log p2-log p1 = ∆Hvap / 2.303 R [1/T2-1/T1]
On putting necessary value
0.5-0.25 = ∆Hvap / 2.303 2 [2.5 10-3 - 5 10-3]
∆Hvap=-0.25× 2.303× 2/2.5× 10-3
∆Hvap = -460 cal mol-1
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Passage IIVariation of log10p with1/Tis given in the following graph (pressure in bar)Q.Enthalpy of vaporisation based on this data isa)460.6 cal mol-1b)-460.6cal mol-1c)200.0cal mol-1d)- 200.0cal mol-1Correct answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Passage IIVariation of log10p with1/Tis given in the following graph (pressure in bar)Q.Enthalpy of vaporisation based on this data isa)460.6 cal mol-1b)-460.6cal mol-1c)200.0cal mol-1d)- 200.0cal mol-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage IIVariation of log10p with1/Tis given in the following graph (pressure in bar)Q.Enthalpy of vaporisation based on this data isa)460.6 cal mol-1b)-460.6cal mol-1c)200.0cal mol-1d)- 200.0cal mol-1Correct answer is option 'A'. Can you explain this answer?.
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