A sample containing 2.50 mol of an ideal gas at 298 K is expanded isothermally a

Question

A sample containing 2.50 mol of an ideal gas at 298 K is expanded isothermally and reversibly from an initial volume of 10.0 L to a final volume of 50.0 L. Calculate G (J) =   and S (J/K) =   for this process. Please use E notation with 2 decimals. For example, -2010.5 is written as -2.01E3. Please note: Even though this is a reversible process, G is not zero. Second law says that G is zero for reversible process only under constant temperature and constant pressure. This process is not constant temperature and constant pressure.

Continue with the last question, but the process is isothermal expansion against a constant external pressure of 0.750 bar (instead of reversible). Calculate G (J) = and S (J) = . please enter your answer in E notation with 2 decimal. For example, -101.456 is written as -1.01E2. Hint: both G and S are state functions.

Explanation / Answer

For Reversible Isothermal Process:

Wrev=-nRT ln(Vfinal/Vinitial) = -9.96E3 J

* for an isothermal process either reversible or irreversible the internal energy change, Delta U, is equal to zero (dU=0)

therefore:

q = -w = 9.96 E3 J

Delta S = qrev / T = 33.45 J/K

If: PiVi = PfVf = nRT

then: Delta(pV) = 0

Delta H = Delta U + Delta(pV) = 0 + 0 = 0

Delta T = 0 (isothermal)

Delta G = Delta H -T Delta S = 9.96 E3 J (which is = q)

isothermal expansion against a constant external pressure of 0.750 bar:

wirrev = -Pext . Delta V = -29.6 J

Delta U = q + w = 0, (q = -w)

Delta H = Delta U + Delta (pV) = 0 + 0 = 0

Delta S = nR ln (Vfinal / Vinitial) = 33.45 J/K

dG = dH - T.dS = - T.dS = 9.96 E3 J

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