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

Q.

The boiling point of a substance X at 1 atm pressure is 500 K. the enthalpy of vapourisation at the boiling point of X (l) is 80 kJ / mol. The molar specific heat of X(l) = 5 × 10-3 mol and Cp of X(g) = 5 × 10-4 kJ/mol.. What is the molar latent heat of vaporization of X(l) at 800 K and at 1 atm?

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

Lesser than 80 kJ

b

Greater than 80 kJ

c

Equal to 80 kJ

d

Since 800K is the higher temperature than the boiling point of  X (l); therefore, enthalpy of vaporization is not applicable.

answer is A.

(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

ΔH1=5×10-3×(500-800)×10-3 kJΔH2=80 kJ;ΔH3=5×10-4×(800-500)×10-3 kJΔH1=-0.0015 kJ;ΔH2=80 kJΔH3=0.00015 kJ;ΔH=ΔH1+ΔH2+ΔH3
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