Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen into a beaker.

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

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Bình An 1 month 2021-08-04T18:58:05+00:00 1 Answers 2 views 0

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    2021-08-04T18:59:50+00:00

    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

           

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