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Physics

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Three moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 430 to 160 K.
(a) What is the work done by (or done to) the gas? Include the algebraic sign.
J

(b) What is the change in the internal energy of the gas? Include the algebraic sign.
J

  • Physics -

    dE = T dS - PdV

    Adiabatic change: Heat absorbed T dS = 0, the change in internal energy is due to work
    P dV performed by the gas.

    Monoatomic gas:

    E = 3/2 N k T

    1 mole = 6.022*10^23

    k = 1.38*10^(-23)J/K

    So, the temperature change gives you know Delta E (which is clearly negative) and that equals minus the work done by the gas (this is positive as work done by the gas comes at the expense of the gas)

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