Q47.The internal energy change (in J) when 90 g of water undergoes complete evaporation at 100°C is .............. (Given : ΔHvap for water at 373 K = 41 kJ / mol, R = 8. 314 JK−1 mol−1 )
What This Question Tests
This question requires calculating the internal energy change during evaporation using the enthalpy of vaporization and the relationship between ΔH and ΔU, considering the change in the number of gaseous moles.
Concepts Tested
Formulas Used
ΔU = ΔH - Δn_g RT
📚 NCERT Sections This Tests
1.34 — Vapour Pressure Of Water At 293 K Is 17.535 Mm Hg. Calculate The Vapour
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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.
1.36 — 100 G Of Liquid A (Molar Mass 140 G Mol–1) Was Dissolved In 1000 G Of Liquid B
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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
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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.
📋 Question Details
- Chapter
- Thermodynamics & Thermochemistry
- Topic
- Internal energy change for phase transition
- Year
- 2020
- Shift
- 02 Sep Shift 1
- Q Number
- Q47
- Type
- Numerical
- NCERT Ref
- Class 11 Chemistry Ch 6: Thermodynamics
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