Explain the expansion is against a vacuum

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Reference no: EM13561000

A fellow student claims that entropy really isn't a state function, for the following reason. Suppose 1.00 mol of an ideal gas expands isothermally and reversibly at 300 K from 10.0 L to 20.0 L. The entropy change for this process is given by ΔS = nR ln (vf/vi) = nR ln 2. Now suppose 1.00 mol of an ideal gas expands isothermally at 300 K against a vacuum from 10.0 L to 20.0 L. Since the expansion is against a vacuum, w = 0; since the gas is ideal and the process is isothermal, ΔE = 0; hence q = 0, and ΔS = 0. Since the gas had the same initial and final states in both experiments, but ΔS had different values, therefore S is not a state function. Is the student correct? If not, identify the error(s) in the student's reasoning.

Reference no: EM13561000

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