Consider a process in which an ideal gas is compressed to one-sixth of its original volume at constant temp?

Calculate the entropy change per mole of gas.

1 Answer

  • If we define the entropy as a function of T and V, then the total differential of the entropy is given by:

    dS = (∂S/∂V)_T dV + (∂S/∂T)_V dT

    For an isothermal process, dT = 0, so:

    dS = (∂S/∂V)_T dV

    Using a Maxwell’s relation (see source) we can write (∂S/∂V)_T = (∂p/∂T)_V, so:

    dS = (∂p/∂T)_V dV

    For an ideal gas, p = n*R*T/V, so (∂p/∂T)_V = n*R/V, and:

    dS = (n*R) * dV/V

    Integrating this, we get:

    ΔS = n*R*ln(V_final/V_initial)

    This is the equation that defines the entropy change for an isothermal expansion of an ideal gas.

    In this case, we have that n = 1 mol, and V_final/V_initial = 1/6, so:

    ΔS = (1mol)*(8.314 J/(mol*K))*ln(1/6)

    ΔS/mol = -14.9 J/(mol*K)

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