Question

Calculate the average volume per molecule for an ideal gas at room temperature and atmospheric pressure. Then take the cube root to get an estimate of the average distance between molecules. How does this distance compare to the size of a molecule

Answer 1

Complete Question

Calculate the average volume per molecule for an ideal gas at room temperature and atmospheric pressure. Then take the cube root to get an estimate of the average distance between molecules. How does this distance compare to the size of a molecule like $N_2$?

Answer:

The  average volume per molecule is  

   $\frac{V}{N} = 4.09 *10^{-26} \ m^3/molecule$

 The average distance between molecules

  $d = 3.45 *10^{-9} \ m$

The size of  $N_2$ is 100 times smaller than the obtained value  

Explanation:

From the question we can deduce that we are considering an ideal

Generally the ideal gas equation is mathematically represented as

       $PV = NkT$

Here  T  is the room temperature with value  T  =  300  \ K  

     k is the Boltzmann constant with value  $k = 1.38 *10^{-23} \ J/K$  

      P  is the atmospheric pressure with value  $P = 1.0 *10^{5} \ N/m^2$

     N is the number of molecules

Now  the  volume per molecule is mathematically deduced from the above equation as

        $\frac{V}{N} = \frac{kT}{P}$

=>      $\frac{V}{N} = \frac{ 1.381 *10^{-23} * 300}{ 1.0*10^{5}}$

=>      $\frac{V}{N} = 4.09 *10^{-26} \ m^3/molecule$

Now the distance is mathematically evaluated as

      $d = \sqrt{\frac{V}{N} }$

=>    $d = \sqrt[3]{4.09*10^{-26}}$

=>     $d = 3.45 *10^{-9} \ m$

Generally the size of  $N_2$  is  115 pm which is  100 times smaller  than the  obtained value  

  Generally the size of  $H_2O$ is  $95.84 \ pm$ which is  10 times smaller than the obtained value