To determine the difference in entropy at 1 atmosphere (ATM) pressure between 1 gram of water at 100°C and 1 gram of steam at 100°C, we need to consider the change in entropy during the phase transition from liquid to vapor.
The change in entropy during a phase transition can be calculated using the equation:
ΔS = Q / T
where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature.
At 100°C, water undergoes a phase transition from liquid to vapor (steam) while maintaining the same temperature. During this transition, energy in the form of heat, known as latent heat, is absorbed by the water molecules to break the intermolecular forces and allow for the phase change.
The heat required for the phase transition is given by the equation:
Q = m * L
where Q is the heat, m is the mass of the substance, and L is the latent heat of vaporization.
For water, the latent heat of vaporization is approximately 540 calories per gram (cal/g).
Let's calculate the change in entropy:
For water at 100°C:
ΔS_water = Q_water / T
ΔS_water = (m * L) / T
ΔS_water = (1 g) * (540 cal/g) / (100°C + 273.15 K)
ΔS_water ≈ 1.989 cal/K
For steam at 100°C:
ΔS_steam = Q_steam / T
ΔS_steam = (m * L) / T
ΔS_steam = (1 g) * (540 cal/g) / (100°C + 273.15 K)
ΔS_steam ≈ 1.989 cal/K
The difference in entropy between 1 gram of water at 100°C and 1 gram of steam at 100°C at 1 ATM pressure is approximately zero. This means that at the same temperature and pressure, the entropy of water and steam is essentially the same.