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

Question

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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Thành Công 2 months 2021-07-28T14:45:38+00:00 1 Answers 0 views 0

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    2021-07-28T14:47:02+00:00

    Answer:

    3.59\times 10^{-21} J

    Explanation:

    We are given that

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

    Universal gas constant,R=8.31451 J/K

    Number of moles,n=3

    Volume ,V=6.1 L=6.1\times 10^{-3}m^3

    Pressure,P=7 atm=7\times 101325 Pa

    PV=nRT

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

    T=173.45 K

    Average kinetic energy=\frac{3}{2}k_BT

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

    Average kinetic energy=3.59\times 10^{-21} J

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