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sdas [7]
3 years ago
6

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
Chemistry
1 answer:
Tatiana [17]3 years ago
4 0

Answer:

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.

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14 - .2x + 1x = 28
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</span>
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