A container with volume 1.83 L is initially evacuated. Then it is filled with 0.246 g of N2. Assume

A container with volume 1.83 L is initially evacuated. Then it is filled with 0.246 g of N2. Assume that the pressure of the gas is low enough for the gas to obey the ideal-gas law to a high degree of accuracy. If the root-mean-square speed of the gas molecules is 192 m/s, what is the pressure of the gas?

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  1. The  pressure is  [tex]P = 1652 \ Pa[/tex]

    Explanation:

    From the question we are told that

        The  volume of the container is  [tex]V = 1.83 \ L = 1.83 *10^{-3 } \ m^3[/tex]

         The mass of  [tex]N_2[/tex] is  [tex]m_n = 0.246 \ g = 0.246 *10^{-3} \ kg[/tex]

         The root-mean-square velocity is  [tex]v = 192 \ m/s[/tex]

    The  root -mean square velocity is mathematically represented as

          [tex]v = \sqrt{ \frac{3 RT}{M_n } }[/tex]

    Now the ideal gas law is mathematically represented as

          [tex]PV = nRT[/tex]

    =>   [tex]RT = \frac{PV}{n }[/tex]

    Where n is the number of moles which is mathematically represented as

             [tex]n = \frac{ m_n }{M }[/tex]

    Where  M  is the molar mass of  [tex]N_2[/tex]

    So  

            [tex]RT = \frac{PVM_n }{m _n }[/tex]

    =>    [tex]v = \sqrt{ \frac{3 \frac{P* V * M_n }{m_n } }{M_n } }[/tex]

    =>    [tex]v = \sqrt{ \frac{ 3 * P* V }{m_n } } }[/tex]

    =>   [tex]P = \frac{v^2 * m_n}{3 * V }[/tex]

    substituting values

        =>    [tex]P = \frac{( 192)^2 * 0.246 *10^{-3}}{3 * 1.83 *10^{-3} }[/tex]

    =>         [tex]P = 1652 \ Pa[/tex]

           

         

  2. Sorry my friend i don't know the answer. i am just in 6th
    [tex]Me : ) it doesn't need to be a paragraph just a list[/tex]

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