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Two vessels are labeled A and B. Vessel A contains NH3 gas at 76°C, and vessel B contains Ne gas at the same temperature. If the average kinetic energy of NH3 is 7.1 × 10−21 J/molecule at 70°C, calculate the root-mean-square speed of Ne atoms in m/s.

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obasola1

Answer:

[tex]\pi _{rms}=656.77m/s

Explanation:

the following formula can be used for deriving the root mean square velocity

[tex]\pi _{rms} =\sqrt({3RT)/M}[/tex]

[tex]\pi^2 _{rms} =({3RT)/M}[/tex]

For te Ne as

T=76°C=76+273=349K

molecular mass of Ne=20.18g/mol=0.02018kg/mol

R=8.314J/K.mol   ..the gas constant

using the formula

[tex]\pi^2 _{rms} =({3*8.314*349)/0.02018}[/tex]

431355.69 j/kg

J=1kg[tex]\frac{kg*m^2}{s^2}[/tex]/(kg)

[tex]\pi^2 _{rms}=m^2/s^2

[tex]\pi _{rms}=m/s

therefore, taking the square root of 431355.69 j/kg

[tex]\pi _{rms}=656.77m/s

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