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

I can't think of a simple way to do this... Are you in thermo? We could use maxwell's equations:

dS = (∂S/∂V) dV + (∂S/∂T) dT

At constant T, we can assume we have the relation:

ds = (∂P/∂V) dV

Using the ideal gas law, we obtain:

P = nRT / V ---> (∂P/∂V)_T = nR / V

∫ ds = ∫ (nR/V) dv

1st order DE:

http://www.wolframalpha.com/input/?i=integrate+dS+...

Vf = Ce^(S/nR)

Vf/C = e^(S/nR)

ln(V / C) = S/nR

Since we want the change in entropy, ∆S makes C = Vi, as the integral would integrate to the final volume:

ln(Vf / Vi) * nR = ∆S

∆S = (1 mol) * (8.314 J / *K * mol)) * (ln(1/2)) = -5.76

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N3XV$81, we can't make that assumption. Remember that a Joule is equal to a Pas * m^3 or really the volume multiplied by pressure. If we half volume, we also double pressure, giving us:

Sf = (8.314 / k mol) * (1/2 m^3) * (2 pas) = 8.314 J / mol K, not 4.157.

∆S = 0 in that case, which is not true.

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Where did you get .5mol(1K), we have the same number of moles start to finish assuming it is a sealed container. T is also the same, where did the .5 come from?

Answer : delta S = Sf - Si

delta S = 4.157J/Kmol - 8.314J/Kmol

delta S = -4.157J/Kmol

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I got my initial entropy, Si, as or equal to R from the gas law.

therefore, the equation I should use for Sf should be:

R = PV/nT

T is constant and I assumed to be = 1K

P = 2

V = 0.5L

n = 0.5mol

R = 8.314J/Kmol

Sf = 8.314J/Kmol = 2P * 0.5L /0.5mol(1K)

Sf = 8.314J/Kmol = 1/0.5

Sf =8.314J/Kmol = 2

Sf = 8.314J/Kmol/2 = 4.157J/Kmol

deltaS = Sf - Si

deltaS = 4.157J/Kmol - 8.314J/Kmol = -4.157J/Kmol