The critical temperature `(T_(c))` and pressure `(P_(c))` of `NO` are `177 K` and `6.48MPa`, respectively, and that of `CCl_(4)` are `550 K` and `4.56MPa`, respectively. Which gas `(a)` has the smaller value for the van der Waals constant `b`, `(b)` has the smaller value for constant `a`, `(c)` has the larger critical volume, and `(d)` is most nearly ideal in behaviour at `300 K` and `1.013MPa`.

Explanation

The given values are
`{:(T_(c)(NO)=177K, T_(c)(C Cl_(4))=550 K), (P_(c)(NO)=6.485MPa, P_(c)(C Cl_(4))=4.56MPa):}`
`a` Since `(P_(c))/(T_(c))=(a//27b^(2))/(8a//27Rb)=(R )/(8b)`,
`b=(T_(c)R)/(8P_(c))`
Thus,
`b(NO)=((177K)(8.314MPadm^(3)K^(-1)mol^(-1)))/((8)(6.485MPa))`
`=28.36 cm^(3)mol^(-1)`
and
`c(C Cl_(4))=((550 K)(8.314MPa cm^(3)K^(-1)mol^(-1)))/((8)(4.56MPa))`
`=125.35 cm^(3)mol^(-1)`
Hence, `b(NO)ltb(C Cl_(4))`.
`b` Since `a=27P_(c)b^(2)`,
`a(NO)=(27)(6.485MPa)(28.36 cm^(3)mol^(-1))^(2)`
`=140.827 kPa dm^(6) mol^(-2)`
`a(C Cl_(4))=(27)(4.56MPa)(125.35 cm^(3)mol^(-1))^(2)`
`=1934.538kPa dm^(6) mol^(-2)`
Hence, `a(NO)lta(C Cl_(4))`.
`c` Since `V_(C)=3b`,
`V_(C)(NO)=3xx(28.36 cm^(3)mol^(-1))`
`=85.08 cm^(3) mol^(-1)`
`V_(C)(C Cl_(4))=3xx(125.35 cm^(3)mol^(-1))`
`=376.05 cm^(3) mol^(-1)`
Hence, `V_(C)(NO)ltV_(C)(C Cl_(4))`.
`d` `NO` is more ideal in behaviour at `300 K` and `1.013 MPa` because its critical temperature is less than `300 K`, whereas the corresponding critical temperature of `C Cl_(4)` is greater than `300 K`.