When the partition is removed, the gas expands and fills both compartments. How many moles n of gas were initially contained in the lower compartment if the entropy change of the gas in this free-expansion process is 17.28 J/K

Question

Chemistry

Efrain Barker

26 March, 03:27

When the partition is removed, the gas expands and fills both compartments. How many moles n of gas were initially contained in the lower compartment if the entropy change of the gas in this free-expansion process is 17.28 J/K

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Answers (1)

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Corinne Potter · Answer 1

n = 1.89 mol.

Explanation:

Think of this as adiabatic free expansion. This means constant temperature - - we know this because the internal energy does not change.

Since the temperature is constant, the entropy change can be given by:

Entropy = nRln (V2/V1)

Entropy = 17.28 J/K. R = 8.314 (gas constant). V2 = 3V. V1 = V.

Thus, we get:

17.28 = n (8.314) ln3

Solve for n. We get n = 1.89 mol.