Question

One mole of carbon dioxide was found to occupy a volume of $1.32litre$ at ${48}^{\circ }C$ under a pressure of $16.4atm$. Calculate the pressure that would have been expected from:
(i) the ideal gas equation,
(ii) van der Waal's equation.
$(a=3.59atmlitr{e}^{2}mo{l}^{-1};b=4.27\times {10}^{-3}litremo{l}^{-1}$ and $R=0.0821litre-atm{K}^{-1}mo{l}^{-1})$.

• A. (i) $10.96atm$ (ii) $18.64atm$.
• B. (i) $14.96atm$ (ii) $18.64atm$.
• C. (i) $19.96atm$ (ii) $17.04atm$.
• D. none

Answer 1

Answer: (C)

(i) $19.96atm$ (ii) $17.04atm$.

$\begin{array}{c}\text{Ideal gas equation}PV=nRT\\ \text{Given}n=1,V=1.32,T=48+273=321K\\ R=0.082,a=3.59,b=4.27\times {10}^{-3}\\ \\ P=\frac{0.0821\times 321}{1.32}=19.965\text{atm}\end{array}$

Vander wall's Equation $\begin{array}{c}(P+\frac{a{n}^2}{V^2})(v-nb)=nRT\\ [P+\frac{3.59}{1.74})(1.32-\frac{4.27}{1000})=0.082\times 321\\ (P+2.06)=\frac{26.32}{1.31}\\ P=17.04\text{atm}\end{array}$

Hence option C is correct