Q53.When ΔHvap = 30 kJ/mol and ΔSvap = 75 J mol−1 K−1 , then the temperature of vapour, at one atmosphere is ______ K.
What This Question Tests
This numerical question directly applies the concept of Gibbs free energy change at equilibrium (ΔG = 0) for a phase transition to calculate the boiling temperature.
Concepts Tested
Formulas Used
ΔG = ΔH - TΔS
ΔG = 0 at equilibrium
📚 NCERT Sections This Tests
1.34 — Vapour Pressure Of Water At 293 K Is 17.535 Mm Hg. Calculate The Vapour
Chemistry Class 11 · Chapter 1
1.34 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
3.23 — The Rate Constant For The Decomposition Of Hydrocarbons Is 2.418 × 10–5S–1
Chemistry Class 11 · Chapter 3
3.23 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
1.36 — 100 G Of Liquid A (Molar Mass 140 G Mol–1) Was Dissolved In 1000 G Of Liquid B
Chemistry Class 11 · Chapter 1
1.36 100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr. 29 Solutions Reprint 2025-26
📋 Question Details
- Chapter
- Thermodynamics & Thermochemistry
- Topic
- Phase transitions and equilibrium
- Year
- 2024
- Shift
- 09 Apr Shift 2
- Q Number
- Q53
- Type
- Numerical
- NCERT Ref
- Class 11 Chemistry Ch 6: Thermodynamics & Thermochemistry
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