Question

Calculate the pressure exerted by one mole of $C{O}_2$ gas at 273 K if the van der Waal's constant, $a=3.592d{m}^{6}atmmo{l}^{-2}$. Assume that the volume occupied by $C{O}_2$ molecules is negligible.

• A. $0.9922atm$
  
• B. $1.9922atm$
• C. $9.9222atm$
• D. $0.1922atm$

Answer 1

The correct option is A $0.9922atm$


Van der Waals' equation for one mole of a gas is
$[P+\frac{a}{V^2}](V-b)=RT.....\left(1\right)$
Given that volume occupied by $C{O}_2$ molecules, 'b' = 0 Hence, (1) becomes $[P+\frac{a}{V^2}]V=RTorP=\frac{RT}{V}-\frac{a}{V^2}$
Using R = 0.082, T = 273K, V =22.4 L for one mole of an ideal gas at 1 atm pressure.
$\therefore P=\frac{0.082\times 273}{22.4}-\frac{3.592}{(22.4{)}^2}=0.9922atm.$