Question

Given,${\text{If }}\dfrac{{\text{P}}}{{{{\text{P}}_{\text{c}}}}}{\text{ = }}{{\text{P}}_{\text{r}}}{\text{,}}\dfrac{{\text{T}}}{{{{\text{T}}_{\text{c}}}}}{\text{ = }}{{\text{T}}_{{\text{r,}}}}{\text{and}}\dfrac{{{{\text{V}}_{\text{m}}}}}{{{{\text{V}}_{{\text{m,c}}}}}}{\text{ = }}{{\text{V}}_{\text{r}}}$ Where
${{\text{P}}_{\text{r}}}$ is reduced pressure ${{\text{P}}_{\text{c}}}$ is critical pressure.
${{\text{T}}_{\text{r}}}$is reduced temperature ${{\text{T}}_{\text{c}}}$is critical temperature.
${{\text{V}}_{\text{r}}}$is reduced volume ${{\text{V}}_{\text{c}}}$is critical volume.
Then the equation of state (or van der Waals equation), only in terms of ${{\text{P}}_{\text{r}}}{\text{,}}{{\text{T}}_{\text{r}}}{\text{and }}{{\text{V}}_{\text{r}}}$is

${\text{A) }}\left( {{{\text{P}}_{\text{r}}}{\text{ + }}\dfrac{{\text{3}}}{{{{\text{V}}_{\text{r}}}^{\text{2}}}}} \right){\text{ = 8 }}{{\text{T}}_{\text{r}}}$
${\text{B)}}\left( {{{\text{P}}_{\text{r}}}{\text{ + }}\dfrac{{\text{3}}}{{{{\text{V}}_{\text{r}}}^{\text{2}}}}} \right){\text{ }}\left( {{\text{3}}{{\text{V}}_{\text{r}}}{\text{ - 1}}} \right)$
${\text{C)}}\left( {{{\text{P}}_{\text{r}}}{\text{ + }}\dfrac{{\text{3}}}{{{{\text{V}}_{\text{r}}}^{\text{2}}}}} \right)\left( {{\text{3}}{{\text{V}}_{\text{r}}}{\text{ - 1}}} \right){\text{ = 4 }}{{\text{T}}_{\text{r}}}$
${\text{D)}}\left( {{{\text{P}}_{\text{r}}}{\text{ + }}\dfrac{{\text{3}}}{{{{\text{V}}_{\text{r}}}^{\text{2}}}}{\text{ }}} \right)\left( {{\text{3}}{{\text{V}}_{\text{r}}}{\text{ - 1}}} \right){\text{ = 8 }}{{\text{T}}_{\text{r}}}$

Answer 1

All the gases are not ideal in nature. Depending on the condition of the gas it behaves as ideal gas. There are three units for measuring the temperature. There are degrees Celsius, kelvin and Fahrenheit. The moles are one of the main units in chemistry. The moles of the molecule depend on the mass of the molecule and molecular mass of the molecule. Chemical reactions are measured by moles only.
Formula used:
The ideal gas equation depends on the pressure, temperature, number of moles, volume of the gas molecules in ideal condition.
The ideal gas equation is,
${\text{PV = nRT}}$
Here, the pressure of the gas is P.
The volume of the gas is V.
The temperature of the gas in kelvin is T.
Gas constant is R.
The number of moles of the Gas molecules is n.

Complete answer:
${\text{If }}\dfrac{{\text{P}}}{{{{\text{P}}_{\text{c}}}}}{\text{ = }}{{\text{P}}_{\text{r}}}{\text{,}}\dfrac{{\text{T}}}{{{{\text{T}}_{\text{c}}}}}{\text{ = }}{{\text{T}}_{{\text{r,}}}}{\text{and}}\dfrac{{{{\text{V}}_{\text{m}}}}}{{{{\text{V}}_{{\text{m,c\;}}}}}}{\text{ = }}{{\text{V}}_{\text{r}}}$ Where
${{\text{P}}_{\text{r}}}$ is reduced pressure ${{\text{P}}_{\text{c}}}$ is critical pressure.
${{\text{T}}_{\text{r}}}$ is reduced temperature ${{\text{T}}_{\text{c}}}$ is critical temperature.
${{\text{V}}_{\text{r}}}$ is reduced volume ${{\text{V}}_{\text{c}}}$ is critical volume.
Then the equation of state (or van der Waals equation), only in terms of ${{\text{P}}_{\text{r}}}{\text{,}}{{\text{T}}_{\text{r}}}{\text{and }}{{\text{V}}_{\text{r}}}$ is
The ideal gas equation is,
${\text{PV = nRT}}$
We change the ideal gas equation to Van der Waals equation is,
${\text{(P + }}\dfrac{{{\text{a}}{{\text{n}}^{\text{2}}}}}{{{{\text{V}}^{\text{2}}}}}{\text{)(V - nb) = nRT}}$
The number of moles of the Gas molecules is ${\text{n = 1}}$.
Hence,
${\text{(P + }}\dfrac{{\text{a}}}{{{{\text{V}}^{\text{2}}}}}{\text{)(V - b) = RT}}$ is Van der Waals equation.
In Van der Waals equation is,
We apply below values in this equation.
${\text{P = }}{{\text{P}}_{\text{c}}}{{\text{P}}_{\text{r}}}$
${\text{T = }}{{\text{T}}_{\text{c}}}{{\text{T}}_{\text{r}}}$
${\text{V = }}{{\text{V}}_{\text{c}}}{{\text{V}}_{\text{r}}}$
${\text{(P + }}\dfrac{{\text{a}}}{{{{\text{V}}^{\text{2}}}}}{\text{)(V - b) = RT}}$
${\text{(}}{{\text{P}}_{\text{c}}}{{\text{P}}_{\text{r}}}{\text{ + }}\dfrac{{\text{a}}}{{{{{\text{(}}{{\text{V}}_{\text{c}}}{{\text{V}}_{\text{r}}})}^{\text{2}}}}}{\text{)(}}{{\text{V}}_{\text{c}}}{{\text{V}}_{\text{r}}}{\text{ - b) = R}}{{\text{T}}_{\text{c}}}{{\text{T}}_{\text{r}}}$
From the Van der Waals equation is we got,
${{\text{P}}_{\text{c}}}{\text{ = }}\dfrac{{\text{a}}}{{{\text{27}}{{\text{b}}^{\text{2}}}}}$
${{\text{V}}_{\text{c}}}{\text{ = 3b}}$
${{\text{T}}_{\text{c}}}{\text{ = }}\dfrac{{{\text{8a}}}}{{{\text{27bR}}}}$
Substitute above three values in Van der Waals equation,
${\text{(}}{{\text{P}}_{\text{c}}}{{\text{P}}_{\text{r}}}{\text{ + }}\dfrac{{\text{a}}}{{{{{\text{(}}{{\text{V}}_{\text{c}}}{{\text{V}}_{\text{r}}})}^{\text{2}}}}}{\text{)(}}{{\text{V}}_{\text{c}}}{{\text{V}}_{\text{r}}}{\text{ - b) = R}}{{\text{T}}_{\text{c}}}{{\text{T}}_{\text{r}}}$
${\text{(}}\dfrac{{\text{a}}}{{{\text{27}}{{\text{b}}^{\text{2}}}}}{{\text{P}}_{\text{r}}}{\text{ + }}\dfrac{{\text{a}}}{{{{{\text{(3b}}{{\text{V}}_{\text{r}}})}^{\text{2}}}}}{\text{)(3b}}{{\text{V}}_{\text{r}}}{\text{ - b) = R}}\dfrac{{{\text{8a}}}}{{{\text{27bR}}}}{{\text{T}}_{\text{r}}}$
${\text{(}}{{\text{P}}_{\text{r}}}{\text{ + }}\dfrac{{\text{3}}}{{{{\text{V}}_{\text{r}}}^{\text{2}}}}{\text{)(3}}{{\text{V}}_{\text{r}}}{\text{ - 1) = 8}}{{\text{T}}_{\text{r}}}$
From the above calculation we conclude the equation of state (or van der Waals equation), only in terms of ${{\text{P}}_{\text{r}}}{\text{,}}{{\text{T}}_{\text{r}}}{\text{and }}{{\text{V}}_{\text{r}}}$is
${\text{(}}{{\text{P}}_{\text{r}}}{\text{ + }}\dfrac{{\text{3}}}{{{{\text{V}}_{\text{r}}}^{\text{2}}}}{\text{)(3}}{{\text{V}}_{\text{r}}}{\text{ - 1) = 8}}{{\text{T}}_{\text{r}}}$

Hence. Option D is correct.

Note:
${\text{(P + }}\dfrac{{\text{a}}}{{{{\text{V}}^{\text{2}}}}}{\text{)(V - b) = RT}}$ is Van der Waal’s equation.
We shift the equation in one side,
${\text{PV + }}\dfrac{{\text{a}}}{{\text{V}}}{\text{ - Pb - }}\dfrac{{{\text{ab}}}}{{{{\text{V}}^{\text{2}}}}}{\text{ = RT}}$
${\text{PV + }}\dfrac{{\text{a}}}{{\text{V}}}{\text{ - Pb - }}\dfrac{{{\text{ab}}}}{{{{\text{V}}^{\text{2}}}}}{\text{ - RT = 0}}$
Multiply the above equation by $\dfrac{{{{\text{V}}^{\text{2}}}}}{{\text{P}}}$
$\dfrac{{{{\text{V}}^{\text{2}}}}}{{\text{P}}}{\text{(PV + }}\dfrac{{\text{a}}}{{\text{V}}}{\text{ - Pb - }}\dfrac{{{\text{ab}}}}{{{{\text{V}}^{\text{2}}}}}{\text{ - RT) = 0}}$
${\text{(}}{{\text{V}}^{\text{3}}}{\text{ + }}\dfrac{{{\text{aV}}}}{{\text{P}}}{\text{ - b}}{{\text{V}}^{\text{2}}}{\text{ - }}\dfrac{{{\text{ab}}}}{{\text{P}}}{\text{ - }}\dfrac{{{\text{RT}}{{\text{V}}^{\text{2}}}}}{{\text{P}}}{\text{) = 0}}$
In Van der Waals equation, we apply below values in this equation.
${\text{P = }}{{\text{P}}_{\text{c}}}$, ${\text{T = }}{{\text{T}}_{\text{c}}}$ and ${\text{V = }}{{\text{V}}_{\text{c}}}$
${\text{V = }}{{\text{V}}_{\text{c}}}$
${\text{V - }}{{\text{V}}_{\text{c}}} = 0$
${{\text{(V - }}{{\text{V}}_{\text{c}}})^3} = 0$
${{\text{V}}^3}{\text{ - 3}}{{\text{V}}_{\text{c}}}{{\text{V}}^2} + 3{\text{V}}{{\text{V}}_{\text{c}}}^2{\text{ - }}{{\text{V}}_{\text{c}}}^3 = 0$
We compare equation ${\text{(}}{{\text{V}}^{\text{3}}}{\text{ + }}\dfrac{{{\text{aV}}}}{{\text{P}}}{\text{ - b}}{{\text{V}}^{\text{2}}}{\text{ - }}\dfrac{{{\text{ab}}}}{{\text{P}}}{\text{ - }}\dfrac{{{\text{RT}}{{\text{V}}^{\text{2}}}}}{{\text{P}}}{\text{) = 0}}$ and ${{\text{V}}^3}{\text{ - 3}}{{\text{V}}_{\text{c}}}{{\text{V}}^2} + 3{\text{V}}{{\text{V}}_{\text{c}}}^2{\text{ - }}{{\text{V}}_{\text{c}}}^3 = 0$ is
${\text{3}}{{\text{V}}_{\text{c}}} = {\text{b + }}\dfrac{{{\text{R}}{{\text{T}}_{\text{c}}}}}{{{{\text{P}}_{\text{c}}}}}$
$3{{\text{V}}_{\text{c}}}^2 = \dfrac{{\text{a}}}{{{{\text{P}}_{\text{c}}}}}$
${{\text{V}}_{\text{c}}}^3 = \dfrac{{{\text{ab}}}}{{{{\text{P}}_{\text{c}}}}}$
By using this relation,
$\dfrac{{{{\text{V}}_{\text{c}}}^{\text{3}}}}{{{\text{3}}{{\text{V}}_{\text{c}}}^{\text{2}}}}{\text{ = }}\dfrac{{{\text{a/}}{{\text{P}}_{\text{c}}}}}{{{\text{ab/}}{{\text{P}}_{\text{c}}}}}$
${{\text{V}}_{\text{c}}}{\text{ = 3b}}$
$3{{\text{V}}_{\text{c}}}^2 = \dfrac{{\text{a}}}{{{{\text{P}}_{\text{c}}}}}$
${{\text{V}}_{\text{c}}}{\text{ = 3b}}$
$3{({\text{3b)}}^2} = \dfrac{{\text{a}}}{{{{\text{P}}_{\text{c}}}}}$
${\text{27}}{{\text{b}}^2} = \dfrac{{\text{a}}}{{{{\text{P}}_{\text{c}}}}}$
${{\text{P}}_{\text{c}}} = \dfrac{{\text{a}}}{{{\text{27}}{{\text{b}}^2}}}$
${\text{3}}{{\text{V}}_{\text{c}}} = {\text{b + }}\dfrac{{{\text{R}}{{\text{T}}_{\text{c}}}}}{{{{\text{P}}_{\text{c}}}}}$ we get ${{\text{T}}_{\text{c}}}{\text{ = }}\dfrac{{{\text{8a}}}}{{{\text{27bR}}}}$.
From this comparison we got values of
${{\text{P}}_{\text{c}}}{\text{ = }}\dfrac{{\text{a}}}{{{\text{27}}{{\text{b}}^{\text{2}}}}}$, ${{\text{V}}_{\text{c}}}{\text{ = 3b}}$ and${{\text{T}}_{\text{c}}}{\text{ = }}\dfrac{{{\text{8a}}}}{{{\text{27bR}}}}$.