Question

The critical temperature and critical pressure of a gas are \[{{31}^{o}}C\]and $728$ atm respectively. Calculate the constants 'a' and 'b'.

Answer 1

Hint: Write down the formula to calculate volume of 1 molecule of gas. We assume the gas to be spherical in nature. Now in one mole of the gas there will be Avogadro number of molecules of gas. So multiply the volume of one gas molecule with Avogadro number to obtain the total volume and thus find radius of the gas molecule.

Complete step by step answer:
Critical temperature is defined as the temperature of a substance in its critical state beyond which it cannot be liquefied. Critical pressure of a fluid is defined as the vapor pressure of the fluid at the critical temperature.
Here we have pre-defined set of formula for constants given by; $a=\dfrac{27{{R}^{2}}{{T}_{c}}^{2}}{69{{P}_{c}}}$ and $b=\dfrac{R\times {{T}_{c}}}{8{{P}_{c}}}$
Critical volume is the volume occupied by the fluid or substance in its critical state (Critical pressure and temperature). Under critical conditions we will define the value of pressure, volume and temperature considering \[n=1.\]
Here we have given data, ${{P}_{c}}=728atm$ and ${{T}_{c}}={{31}^{o}}C=304K$ and $R=0.082$ here R is constant.
Thus, by substituting the given values in constant’s formula we get their respective values for given dat;
$a=\dfrac{27{{R}^{2}}{{T}_{c}}^{2}}{69{{P}_{c}}}=\dfrac{27\times {{0.082}^{2}}\times {{304}^{2}}}{64\times 728}=0.36{{L}^{2}}atm/mo{{l}^{2}}$
$b=\dfrac{R\times {{T}_{c}}}{8{{P}_{c}}}=\dfrac{0.082\times 304}{8\times 728}=0.0043L/mol$

 Note: Compressibility factor (Z) is a correction factor which describes the deviation of a real gas from its behavior of an ideal gas. It is simply the ratio of molar volume of the gas to the molar volume of an ideal gas subjected to the same identical temperature and pressure. - For ideal gases\[Z=1,PV=nRT\]