Question

The critical temperature and pressure of ${\text{C}}{{\text{O}}_2}$ gas are $304.2{\text{ K}}$ and $72.9{\text{ atm}}$ respectively. What is the radius of ${\text{C}}{{\text{O}}_2}$ molecules assuming it to behave as Van Der Waals gas?

Answer 1

Here, we are assuming that the carbon dioxide gas behaves as Van Der Waals gas so for van der waals gas, Critical temperature ${{\text{T}}_{\text{c}}} = \dfrac{{{\text{8a}}}}{{{\text{27bR}}}}$ - (i), and Critical pressure ${{\text{P}}_{\text{c}}} = \dfrac{{\text{a}}}{{{\text{27}}{{\text{b}}^{\text{2}}}}}$ - (ii) where the symbols have usual meaning, and b is the excluded volume per mole of gas and a is the reduced pressure. From, eq. (i) and (ii), form a formula for b and solve it by putting the given values. Then use the formula of the excluded volume per mole of gas which can be expressed as-

\Rightarrow {\text{b = 4}}{{\text{N}}_0}\left\\{ {\dfrac{4}{3}\pi {r^3}} \right\\} where ${{\text{N}}_{\text{0}}}$ is Avogadro number, r= radius

Then, put the given values and solve it to get the answer.

Complete step-by-step answer:

Given, critical temperature of carbon dioxide= $304.2{\text{ K}}$

The critical pressure of carbon dioxide gas= $72.9{\text{ atm}}$

We have to find the radius of carbon dioxide gas. Here we have to assume that the gas behaves as Van Der Waals gas.

So we know that for van der waals gas, Critical temperature ${{\text{T}}_{\text{c}}} = \dfrac{{{\text{8a}}}}{{{\text{27bR}}}}$ - (i),

Critical pressure ${{\text{P}}_{\text{c}}} = \dfrac{{\text{a}}}{{{\text{27}}{{\text{b}}^{\text{2}}}}}$ - (ii)

From (i) and (ii), we can get the value of b. So, we have-

$\Rightarrow {\text{b}} = \dfrac{{\text{R}}}{8} \times \dfrac{{{{\text{T}}_{\text{c}}}}}{{{{\text{P}}_{\text{c}}}}}$

Now, on putting the given values of critical temperature and critical pressure and taking R= $0.0821$ , we have-

$\Rightarrow b = \dfrac{{0.0821}}{8} \times \dfrac{{304.2}}{{72.9}}$

On solving, we get-

$\Rightarrow b = 0.428{\text{ litremo}}{{\text{l}}^{ - 1}}$

We know that $1{\text{L}} = 1000{\text{c}}{{\text{m}}^{\text{3}}}$

So b= $0.428 \times {10^3}{\text{c}}{{\text{m}}^3}{\text{mo}}{{\text{l}}^{ - 1}}$

Now, we know that b is the excluded volume per mole of gas which can also be expressed as-

\Rightarrow {\text{b = 4}}{{\text{N}}_0}\left\\{ {\dfrac{4}{3}\pi {r^3}} \right\\} where ${{\text{N}}_{\text{0}}}$ is Avogadro number, r= radius

On putting the obtained value of b and Avogadro number, we get-

\Rightarrow 0.428 \times {10^3}{\text{ = 4}} \times {\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{23}}\left\\{ {\dfrac{4}{3}\pi {r^3}} \right\\}

On solving, we get-

\Rightarrow 0.01776523 \times {\text{1}}{{\text{0}}^{ - 20}}{\text{ = }}\left\\{ {\dfrac{4}{3}\pi {r^3}} \right\\}

On rearranging, we get-

$\Rightarrow {r^3}{\text{ = }}0.01776523 \times {\text{1}}{{\text{0}}^{ - 20}} \times \dfrac{3}{{4\pi }}$

On further solving and putting the value of pi constant, we get-

$\Rightarrow {r^3}{\text{ = }}0.013323925 \times {\text{1}}{{\text{0}}^{ - 20}} \times \dfrac{7}{{22}}$

On further solving, we get-

$\Rightarrow {r^3}{\text{ = }}0.0042394306 \times {\text{1}}{{\text{0}}^{ - 20}}$

We can also write it as-

$\Rightarrow r{\text{ = }}\sqrt[3]{{0.042394306 \times {\text{1}}{{\text{0}}^{ - 21}}}}$

On further solving, we get-

$\Rightarrow r = 0.348687072 \times {10^{ - 7}}$ cm

We can also write it as-

$\Rightarrow r = 3.48687072 \times {10^{ - 8}}$

We know that $1{{\text{A}}^\circ } = {10^{ - 8}}{\text{cm}}$ so we get-

$\Rightarrow r = 3.48687072{\text{ }}{{\text{A}}^\circ }$

Hence the radius = $3.48687072{\text{ }}{{\text{A}}^\circ }$.

Note: Here the student may make a mistake if he/she forgets to find the cube root of ${10^{ - 21}}$ and writes it as the same. Here we have converted litre into cm cube so that it will be easier for us to find the cube root of the obtained value. This is also the reason why we change the integral power of ten from $- 20$ to $- 21$ so that we can easily find the cube root of the power.