Question

# A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase ΔT of 20.0∘C; at constant pressure. As a result, the bubble expands. The helium is monoatomic and ideal. How much work W is done by the helium as it expands against the pressure of the surrounding water during the temperature increase? Use the first law of thermodynamics.​

A

820 J

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B

831 J

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C

883 J

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D

805 J

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Solution

## The correct option is B 831 J The bubble displaces some water when it expands, thus doing some work W, which according to the first law of thermodynamics equals - W = Q − ΔEint. Since the expansion process occurs at a constant pressure we have to use the molar specific heat at constant pressure CP, to find the heat gained ΔQ. On the other hand, ΔEint always depends on the molar specific heat at constant volume CV. In kinetic theory, CP and CV are related as - CP − CV = R ⇒ CP = CV + R. Therefore, W = Q − ΔEint = nCPΔT − nCVΔT = n(CP − CV)ΔT = nRΔT = (5 × 8.314 × 20)J = 831.4 J. Note that the work done in a constant-pressure, volume-expansion process is just W = PΔV, which for an ideal gas, is just nRΔT, thus in agreement with the result we obtained from the first law.

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