Question

Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen into a beaker. After the nitrogen evaporates, how much volume does it occupy if its density is equal to that of the dry air at sea level

Answer 1

Answer:

The  value is  $V_n = 2.2498 \ m^3$

Explanation:

From the question we are told that

   The volume of  liquid nitrogen is  $V_n = 3.6 \ L= 3.6 *10^{-3} \ m^3$

   The  density of  nitrogen at gaseous form   is  $\rho_n = 1.2929 \ kg/m^3$  =  The dry air at sea level

   

Generally the density of nitrogen at liquid form is  

         $\rho _l = 808 \ kg/m^3$

And this is mathematically represented as

      $\rho_l = \frac{m}{V_l }$

=>   $m = \rho_l * V_l$

Now the density of  gaseous nitrogen is

       $\rho_n = \frac{m}{V_n }$

=>   $m = \rho_n * V_n$

Given that the mass is constant

       $\rho_n * V_n = \rho_l * V_l$

        $1.2929* V_n = 808 * 3.6*10^{-3}$

=>   $V_n = 2.2498 \ m^3$