Calculate the entropy changes for a 1.35 mol sample of a perfect gas, at 25°C, as it expands from a volume of 3L to 5.5L isothermally both reversibly and against a vacuum.  

The change of entropy of the gas as it expands isothermally either reversibly or against the vacuum is 20.6 J/K.

Expert Answers

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The entropy change for a process is determined by `dS = (dQ)/T` , where `dQ ` is the heat transfer, and T is the temperature.

By definition, the temperature stays constant during an isothermal process. This means that the internal energy of the gas is also constant. Therefore, according to the first law of thermodynamics, the heat transfer during the isothermal process equals the work done by the gas as it expands from initial volume `V_i` to final volume `V_f` . If the process is reversible, the work is given by

`W = nRTln(V_f/V_i)` .

Here, n is the number of moles and `R = 8.31 J/(K*mol)` is the universal gas constant.

The change of entropy during this reversible isothermal process is then

`DeltaS = (Delta Q)/T = W/T = nRln(V_f/V_i)` .

For the process described in the problem, n = 1.35 mole and initial and final volumes are 3L and 5.5L, respectively. Then, the change of entropy is

`Delta S = 1.35*8.31*ln(5.5/3) = 20.6 J/K ` .

If the gas expands isothermally against the vacuum, this process is very fast and therefore not reversible. For this kind of process, the pressure as a function of volume is not defined. However, the change of entropy of the gas is independent of the process. The change of entropy of gas as it expands from the initial to final volume will be the same as for reversible isothermal expansion: 20.6 J/K.

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