wiz-icon
MyQuestionIcon
MyQuestionIcon
13
You visited us 13 times! Enjoying our articles? Unlock Full Access!
Question

Match gases under specified conditions listed in Column 1 with their properties/laws in Column II.

Column IColumn IIAHydrogen gas p = 200 atm,T = 273 KpCompression factor = 1BHydrogen gas p = 0 atm,T = 273 KqAttractive forces are dominantCCO2p = 1 atm,T = 273 KrpV=nRTDReal gas with very large molar volumespVnb=nRT


A

A(r), B(p), C(s), D(q)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

A(p,s), B →(r), C(p,q), D(p,s)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

A(q), B(p), C(s), D(r)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

A(p), B(q), C(r), D(s)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

A(p,s), B →(r), C(p,q), D(p,s)


A (p, s); B (r); C (p, q); D (p, s)

(A) Z = 1 for ideal gas. At low temperature and high pressure, real gases show ideal behaviour which means the equation [p+an2V2][V - nb] = nRT reduces to p(V - nb) = nRT.

(B) For H2 gas, the value of Z = 1 at p = 0 and it increases continuously on increasing p

(C) CO2 molecules have larger attractive forces, under normal conditions.

(D) As Z = pVmRT, at very large molar volume Z is not equal to 1.


flag
Suggest Corrections
thumbs-up
7
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
First Law of Thermodynamics
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon