wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let mixture be prepared by mixing nA moles of liquid A and nB moles of liquid B. Let PA and PB be the partial pressures of two constituents and PA and PB are the vapour pressures in pure state. Which of the following relations are true?

A
PA=PA×ηAηBPB=PB×ηBηA
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
PA=PAηAηA+ηBPB=PBηBηA+ηB
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
PA=PA×ηAηA+ηBPB=PB×ηBηA+ηB
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
PA=PA.nBPB=PB.nB
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C PA=PA×ηAηA+ηBPB=PB×ηBηA+ηB
If two volatile liquids of vapor pressures PA,PB are mixed together to form a solution. The partial pressure of one component is equal to Vapour pressure of pure component multiplied by mole fraction.
So putting this definition in the equations
PA=PA×ηAηA+ηBPB=PB×ηBηA+ηB

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Liquids in Liquids and Raoult's Law
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon