Question

What is the change in entropy of \(10 \mathrm{g}\) of steam at $100^{\circ} \mathrm{C}\( as it condenses to water at \)100^{\circ} \mathrm{C} ?$ By how much does the entropy of the universe increase in this process?

Short answer

Answer: The change in entropy of the steam as it condenses to water is approximately -60.6 J/K. The increase in the entropy of the universe is 0 J/K. Since the increase in the entropy of the universe is zero, the process is an equilibrium process.

Step-by-step solution

Convert the given quantities to the appropriate units

First, convert the mass of steam from grams to kilograms and the temperature from Celsius to Kelvin: - mass, m = 10g = 0.01 kg - temperature, T = 100°C = 373.15 K

Determine the heat of transformation for steam to water

We will use the latent heat of vaporization, Lv for this purpose. Lv for water is approximately 2.26 x 10^6 J/kg.

Calculate the change in entropy of the system (steam)

Using the formula ΔS = m * L / T, we have: ΔS_system = (0.01 kg) * (2.26 x 10^6 J/kg) / (373.15 K) ΔS_system ≈ -60.6 J/K The negative sign indicates that the system loses entropy when steam condenses to water.

Calculate the change in entropy of the surroundings

As heat is transferred from the steam to the surroundings, the entropy change of the surroundings (ΔS_surroundings) is given by: ΔS_surroundings = - ΔQ / T Since the heat is given off by the steam during the transformation process, we have, ΔQ = - m * L, and substituting the values, we get: ΔQ = - (0.01 kg) * (2.26 x 10^6 J/kg) = - 22,600 J Now, we determine ΔS_surroundings: ΔS_surroundings = - (- 22,600 J) / (373.15 K) ΔS_surroundings ≈ 60.6 J/K

Calculate the increase in the entropy of the universe

Now, we find the increase in the entropy of the universe, ΔS_univ, which is the sum of the entropy change of the system and the surroundings: ΔS_univ = ΔS_system + ΔS_surroundings ΔS_univ = -60.6 J/K + 60.6 J/K ΔS_univ = 0 J/K In this process, the increase in the entropy of the universe is zero, which means that the process is an equilibrium process, and it respects the second law of thermodynamics, as ΔS_univ is not negative.