Question

a gas obeys the equation of state p(v-b)=RT.for the gas b=0.0391L/mol.calculate the fugacity coefficient for the gas at 1000°c and 1000atm

Answer 1

Answer:

The fugacity coefficient is  $[\frac{f}{p} ] = 1.45$

Explanation:

From the question we are told that

  The gas obeys the equation $p(v-b) = RT$

   The value  of b is  $b = b = 0.0391 \ L /mol$

   The pressure is  $p = 1000 \ atm$

    The temperature is $T= 1000^oC = 1273 K$

generally

        $RT ln[\frac{f}{p} ] = \int\limits^{p}_{o} [ {v_{r} -v_{i}} ]\, dp$

Where  $\frac{f}{p}$ is the fugacity coefficient

          $v_r$ is the real volume which is mathematically evaluated from above equation  as

           $v_r = \frac{RT}{p} + b$

           $v_r = \frac{RT}{p} + 0.0391$

and     $v_{i}$ is the ideal volume which is evaluated from the ideal gas equation (pv = nRT , at  n= 1) as

         $v_{i} = \frac{RT}{p}$

So

     $RT ln[\frac{f}{p} ] = \int\limits^{1000}_{o} [[ \frac{RT}{p} + 0.0391] - [\frac{RT}{p} ]} ]\, dp$

=>      $RT ln[\frac{f}{p} ] = \int\limits^{1000}_{o} [0.391 ]\, dp$

=>    $RT ln[\frac{f}{p} ] = [0.391p]\left | 1000} \atop {0}} \right.$

=>   $RT ln[\frac{f}{p} ] = 38.1$

   So

         $ln[\frac{f}{p} ] = \frac{39.1}{RT}$

Where R is the gas constant with value $R = 0.082057\ L \cdot atm \cdot mol^{-1}\cdot K^{-1}$

         $[\frac{f}{p} ] = \frac{39.1}{ 2.303 *0.082057 * 1273}$

        $[\frac{f}{p} ] = 1.45$