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

The pressure is
`P = 1652 \ Pa`

Step-by-step 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{ (3 RT)/(M_n ) }`

Now the ideal gas law is mathematically represented as

`PV = nRT`

=>
`RT = (PV)/(n )`

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

`n = ( m_n )/(M )`

Where M is the molar mass of
`N_2`

So

`RT = (PVM_n )/(m _n )`

=>
`v = \sqrt{ (3 (P* V * M_n )/(m_n ) )/(M_n ) }`

=>
`v = \sqrt{ ( 3 * P* V )/(m_n ) } }`

=>
`P = (v^2 * m_n)/(3 * V )`

substituting values

=>
`P = (( 192)^2 * 0.246 *10^(-3))/(3 * 1.83 *10^(-3) )`

=>
`P = 1652 \ Pa`