Question

yas. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV VS. 1N plot is shown below. The value of the van der Waals constant a (atm. litre mol) (Graph not to scale) is (A) 1.0 (C) 1.5 (B) 4.5 (D) 3.0 PV (fixer-atm mol') W (mol liter) (2012) The term that corrects for the attracting

IIT/JEE

Chemistry

Solution

4.0 (1 ratings)

As we know ( :left(P+frac{n^{2} a}{v^{2}}right)(V-n b)=n R T ) for 1 mole ( & ) when ( b=0 ) then ( e q^{u} ) becomes ( r ) [ begin{array}{l} left.f^{p}+frac{a}{v^{2}}right)^{v}=R T quad quad p v=R T-frac{a}{v} end{array} ] As ( y=m x+c )