CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
9
You visited us 9 times! Enjoying our articles? Unlock Full Access!
Question

Given: dE=TdS−pdV and H=E+pV.


Which one of the following relation is true?

A
dH=TdS+Vdp
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
dH=SdT+Vdp
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
dH=SdTVdp
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
dH=dEpdV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A dH=TdS+Vdp
Given,

dE=TdSpdV ...(i)

H=E+pV ...(ii)

Differentiating Eq. (ii),we gte:

dH=dE+pdV+Vdp ...(iii)

From Eq. (i) and (iii), we get-

dH=Tds+Vdp

Hence, option A is correct.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon