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

Match the Van der Waal's for one mole of gas given in column B to the conditions in column A.
Column 1Column 2AAt low pressureiP(Vb)=RTBAt high pressureii(P+aV2)V=RTCAt low pressure and high temperatureiiiPV=RTDAt room temperature and pressureiv(P+aV2)(Vb)=RT

A
A(i);B(ii);C(iii);D(iv)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
A(ii);B(i);C(iii);D(iv)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
A(iv);B(iii);C(ii);D(i)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
A(iii);B(ii);C(i);D(iv)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B A(ii);B(i);C(iii);D(iv)
We know that,
At room temperature and pressure, generally Van der Waal's equation for one mole of gas is
(P+aV2)(Vb)=RT
At low pressure,
PV
V>>b
Neglect b
Van der Waal's equation =(P+aV2)V=RT
At high pressure,
P is very large
P>>aV2
Neglect aV2
Van der Waal's equation =P(Vb)=RT
At low pressure and high temperature, real gas behaves as an ideal gas. So, Van der Waal's equation reduces to
PV=RT
Hence, option (b) is correct.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon