Question

In an isothermal reversible expansion at 27°C, an ideal gas does 20 J of work. What is the entropy change of the gas?

(a) 0 J/K
(b) 0.74 J/K
(c) 1.23 J/K
(d) 2.45 J/K

Answer 1

The entropy change for an isothermal reversible expansion can be calculated by dividing the work done by the absolute temperature. However, the calculation here does not match the multiple-choice answers provided.

Step-by-step explanation:

The entropy change (\(\Delta S\)) during an isothermal process can be calculated using the formula \(\Delta S = \frac{Q}{T}\), where \(Q\) is the heat exchanged and \(T\) is the absolute temperature. Since we are considering a reversible expansion, the work done by the gas (\(W\)) is equal to the heat added, hence \(Q = W = 20\ J\). The temperature (27 \(\degree C\) or 300 K) must be converted to Kelvins by adding 273 to the Celsius temperature.

Using the formula:

\(\Delta S = \frac{Q}{T} = \frac{20\ J}{300\ K}\)

\(\Delta S = 0.0667\ J/K\)

Since this answer is not listed in the options, there might be an understanding or calculation error, or possibly a typo in the question or the options provided. It is important to review the question and make sure that all values provided are accurate before concluding.