Question

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?

Answer 1

Answer:

The  pressure is  $P = 1652 \ Pa$

Explanation:

From the question we are told that

    The  volume of the container is  $V = 1.83 \ L = 1.83 *10^{-3 } \ m^3$

     The mass of  $N_2$ is  $m_n = 0.246 \ g = 0.246 *10^{-3} \ kg$

     The root-mean-square velocity is  $v = 192 \ m/s$

The  root -mean square velocity is mathematically represented as

      $v = \sqrt{ \frac{3 RT}{M_n } }$

Now the ideal gas law is mathematically represented as

      $PV = nRT$

=>   $RT = \frac{PV}{n }$

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

         $n = \frac{ m_n }{M }$

Where  M  is the molar mass of  $N_2$

So  

        $RT = \frac{PVM_n }{m _n }$

=>    $v = \sqrt{ \frac{3 \frac{P* V * M_n }{m_n } }{M_n } }$

=>    $v = \sqrt{ \frac{ 3 * P* V }{m_n } } }$

=>   $P = \frac{v^2 * m_n}{3 * V }$

substituting values

    =>    $P = \frac{( 192)^2 * 0.246 *10^{-3}}{3 * 1.83 *10^{-3} }$

=>         $P = 1652 \ Pa$