Boltzmann’s constant is 1.38066×10⁻²³ J/K, and the universal gas constant is 8.31451 J/K · mol. If 3 mol of a gas is confined to a 6.1 L ves

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Boltzmann’s constant is 1.38066×10⁻²³ J/K, and the universal gas constant is 8.31451 J/K · mol. If 3 mol of a gas is confined to a 6.1 L vessel at a pressure of 7 atm, what is the average kinetic energy of a gas molecule? Answer in units of J.

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  1. Answer:

    [tex]3.59\times 10^{-21} J[/tex]

    Explanation:

    We are given that

    Boltzmann’s constant, [tex]k_B=1.38066\times 10^{-23} J/K[/tex]

    Universal gas constant,R=8.31451 J/K

    Number of moles,n=3

    Volume ,V=6.1 L=[tex]6.1\times 10^{-3}m^3[/tex]

    Pressure,P=7 atm=[tex]7\times 101325 Pa[/tex]

    [tex]PV=nRT[/tex]

    [tex]T=\frac{PV}{nR}=\frac{7\times 101325\times 6.1\times 10^{-3}}{3\times 8.31451}[/tex]

    T=173.45 K

    Average kinetic energy=[tex]\frac{3}{2}k_BT[/tex]

    Average kinetic energy=[tex]\frac{3}{2}(1.38066\times 10^{-23}\times 173.45[/tex]

    Average kinetic energy=[tex]3.59\times 10^{-21} J[/tex]

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