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What is the entropy change when...?

1)

1 cm^3 of ideal gas at standard temperature and pressure expands freely into an equal volume so that the volume at the end of the expansion is twice the starting volume

2)

Two unlike ideal gasses occupying different, but identical, volumes (1 cm^3) at standard temperature and pressure which are initially separated by a partition are allowed to diffuse into each other. THe total volume occupied by each gas after the diffusion. The total volume occupied by each gas after the diffusion is therefore twice the starting volume.

3)

Two equal volumes of the same gas (1cm^3) at standard temperature and pressure are separated from each other by a partition which is removed to create twice the volume at standard temperature and pressure.

1 Answer

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  • 9 years ago
    Favorite Answer

    Tds = du + pdv = dh - vdp

    Δs = cv*ln(T2/T1) + R ln(v2/v1) = cp*ln(T2/T1) - R ln(p2/p1)

    1.

    Δs = cv*ln(T2/T1) + R ln(v2/v1)

    Δs = R ln2 = 5.76282566 J/mol.K

    2.

    Δs = Δs1+Δs2 = R ln(vf/v1) + R*ln(vf/v2)

    Δs = R*ln2+R*ln2

    Δs = 11.5256513 J/mol.K

    3. Twice the volume is final volume for both gasses, hence at the end each gas has the same initial volume.

    Δs = Δs1+Δs2 = R ln(vf/v1) + R*ln(vf/v2)

    Δs = R ln1 + R*ln1

    Δs = 0 J/mol.K

    Δs is zero, because ideal gas state that volume occupy by gas can be neglected. However initial condition has lower volume because matter occupy certain small volume, vf>v2, indicates Δs > 0.

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