Pressure of ideal and real gases at 0 K are

Answer

C

Explanation

`(i) because` ideal gas z=1 =1pV=nRT
or `(pV)/(nRT)=1(Z)`
for real gas z `n` 1 i.e p `lt` 0
(ii) For ideal gas
Temperature must be high and pressure must below i.e
pressure =0 (or negigible )
Temperature `gt` 0 (or high)