Considere 2 mols de um gás monoatômico ideal, inicialmente a 350K. O gás é aquecido reversivelmente até 500K enquanto seu volume é mantido constante. Determine a variação de entropia do gás. Dado: constante dos gases ideais: R cong 8,3J/mol cdot K.

To determine the entropy variation of the ideal gas,<|begin_of_text|>201C, we can use the following equation:ΔS = n * R * ln(T2 / T1)where ΔS is the entropy variation, n is the number of moles, R is the ideal gas constant, T1 is the initial temperature, and T2 is the final temperature.In this case, we have 2 moles of an ideal monoatomic gas, initially at 350 K. The gas is heated reversibly to 500 K, while its volume is kept constant.First, let's find the entropy variation:ΔS = 2 * 8,3 * ln(500 / 350)ΔS = 24,9 J/K

Therefore, the correct answer is alternative C) 24,9 J/K.

This result makes sense, since the entropy of the system increases as the temperature increases.

In an isothermal process, the entropy variation is directly proportional to the logarithm of the temperature ratio. In this case, the temperature ratio is approximately 1.43 (500 K / 350 K), which results in an entropy variation of approximately 24.9 J/K.

It's worth noting that, in a reversible process, the entropy variation of the system is equal to the entropy variation of the surroundings, but with opposite signs. In this case, the entropy of the surroundings decreases by 24.9 J/K, while the entropy of the system increases by the same amount.