A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N 2 ​ . Assume that the pres

Question

A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N
2

. 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 182 m/s, what is the pressure of the gas?

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Huyền Thanh 6 months 2021-07-15T03:29:29+00:00 1 Answers 21 views 0

Answers ( )

    0
    2021-07-15T03:30:47+00:00

    Answer:

    0.015 atm

    Explanation:

    The pressure of the gas can be calculated using Ideal Gas Law:

     p = \frac{nRT}{V}

    Where:

    n: is the number of moles of the gas

    R: is the gas constant = 0.082 L*atm/(K*mol)

    V: is the volume of the container = 1.64 L

    T: is the temperature

    We need to find the number of moles and the temperature. The number of moles is:

     n = \frac{m}{M}

    Where:

    M: is the molar mass of the N₂ = 14.007 g/mol*2 = 28.014 g/mol

    m: is the mass of the gas = 0.226 g

     n = \frac{0.226 g}{28.014 g/mol} = 8.07 \cdot 10^{-3} moles

    Now, the temperature can be found using the following equation:

     v_{rms} = \sqrt{\frac{3RT}{M}}    

    Where:

    R: is the gas constant = 0.082 L*atm/K*mol = 8.314 J/K*mol

     v_{rms}: is the root-mean-square speed of the gas = 182 m/s

    By solving the above equation for T, we have:

    T = \frac{v_{rms}^{2}*M}{3R} = \frac{(182 m/s)^{2}*28.014 \cdot 10^{-3} Kg/mol}{3*8.314 J K^{-1}mol^{-1}} = 37.20 K        

    Finally, we can find the pressure of the gas:

    p = \frac{nRT}{V} = \frac{8.07 \cdot 10^{-3} mol*0.082 L*atm* K^{-1}*mol^{-1}*37.20 K}{1.64 L} = 0.015 atm

    Therefore, the pressure of the gas is 0.015 atm.

    I hope it helps you!

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