QJEEM21Ched89: JEE Main 2021 Chemistry - Class 11 JEEM21Ched89 Solution: A certain gas obeys P(V m b) = RT. The value of is . The value…

For 1 mole of a real gas, the van der Waals' equation is,

$\left( {p + {a \over {V_m^2}}} \right)({V_m} - b) = RT$

At very high pressure, the equation becomes,

$p({V_m} - b) = RT$

$\Rightarrow p{V_m} = RT + pb \Rightarrow {{p{V_m}} \over {RT}} = 1 + {{pb} \over {RT}}$

$\Rightarrow Z = 1 + {{pb} \over {RT}}$ [ $\because$ $Z = {{p{V_m}} \over {RT}}$ = compressibility]

$\therefore$ ${\left( {{{\delta Z} \over {\delta p}}} \right)_T} = 0 + {b \over {RT}} + {b \over {RT}} = {{xb} \over {RT}}$ (Given)

$\Rightarrow x = 1$