Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Assuming that water vapour is an ideal gas, the internal energy change ∆U when 1 mole of water is vaporised at 1 bar pressure and 1000C, (given: molar enthalpy of vaporisation of water = 41 kJ mol-1 at 1 bar and 373 K and R = 8.3 J mol-1 K mol-1) will be

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

4.100 kJ mol-1

b

3.7904 kJ mol-1

c

37.904 kJ mol-1

d

41.00 kJ mol-1

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

H2Ol →vaporisation H2Og ∆ng=1-0=1 ∆H=∆U+∆ngRT ∆U=∆H-∆ngRT =41-8.3 × 10-3×373=37.9 kJ mol-1
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring