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?
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]
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