Question

Under what pressure will carbon dioxide have the density of 2.2 g/liter at 300 K? For $C{O}_2,a=3.6atm{l}^{2}mo{l}^{-2}$ and $b=0.05mo{l}^{-1}l.$

• A. 1.23 atm
• B. 1.28 atm
• C. 2.46 atm
• D. 0.64 atm

Answer 1

The correct option is A 1.23 atm

Solution:- (A) $1.23atm$

From Van der Waal's equation for a real gas-

$(P+\frac{a{n}^2}{V^2})(V-nb)=nRT.....\left(1\right)$

As we know that,

No. of moles $\left(n\right)=\frac{\text{Weight}\left(w\right)}{\text{Mol. weight}\left(M\right)}.....\left(2\right)$

$\because$ Density $\left(\rho \right)=\frac{w}{V}.....\left(3\right)$

From equation $\left(2\right)\&\left(3\right)$, we get

$\frac{n}{V}=\frac{\rho }{M}$

Substituting the value of $n$ in equation $\left(1\right)$, we get

$(P+\frac{a{\rho }^2}{M^2})(1-\frac{\rho }{M}b)=\frac{\rho }{M}RT$

Given:-

$\rho =2.2g/L$

$T=300K$

Molecular weight of $C{O}_2\left(M\right)=44g$

$a=3.6atm{L}^2/{mol}^2$

$b=0.05L/mol$

$P=?$

Therefore,

$(P+\frac{3.6\times {\left(2.2\right)}^2}{{\left(44\right)}^2})(1-\frac{2.2\times 0.05}{44})=\frac{2.2}{44}\times 0.0821\times 300$

$\Rightarrow (P+\frac{3.6}{400})(1-\frac{0.05}{20})=0.0821\times 15$

$\Rightarrow P+\frac{3.6}{400}=\frac{0.0821\times 15\times 20}{19.95}$

$\Rightarrow P=1.235-\frac{3.6}{400}\approx 1.23atm$