Question

One mole of an ideal monoatomic gas (initial temperature = 127°C) is expended adiabatically and reversibly from 5 m³ to 20 m³. Calculate $\Delta S_{sys}$, $\Delta S_{surr}$ and $\Delta S_{univ}$.

Answer 1

$\Delta S_{sys} = 0$, $\Delta S_{surr} = 0$, $\Delta S_{univ} = 0$

Answer 2

The problem asks to calculate the change in entropy for the system ($\Delta S_{sys}$), surroundings ($\Delta S_{surr}$), and universe ($\Delta S_{univ}$) for an ideal monoatomic gas undergoing a reversible adiabatic expansion.

1. Calculate $\Delta S_{sys}$ (Change in entropy of the system): For any reversible process, the change in entropy of the system is given by: $\Delta S_{sys} = \int \frac{dQ_{rev}}{T}$

Since the process is adiabatic, there is no heat exchange, meaning $dQ_{rev} = 0$. Therefore, $\Delta S_{sys} = \int \frac{0}{T} = 0$

2. Calculate $\Delta S_{surr}$ (Change in entropy of the surroundings): For an adiabatic process, there is no heat exchange between the system and the surroundings. This means the heat absorbed by the surroundings ($Q_{surr}$) is zero. The change in entropy of the surroundings is given by: $\Delta S_{surr} = \frac{Q_{surr}}{T_{surr}}$

Since $Q_{surr} = 0$, $\Delta S_{surr} = \frac{0}{T_{surr}} = 0$

3. Calculate $\Delta S_{univ}$ (Change in entropy of the universe): The change in entropy of the universe is the sum of the change in entropy of the system and the surroundings: $\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr}$

Substituting the calculated values: $\Delta S_{univ} = 0 + 0 = 0$

This result is consistent with the definition of a reversible process, for which the total entropy change of the universe is always zero. The initial temperature, volumes, and the type of gas (monoatomic) are additional information that would be necessary if the process were not explicitly stated as reversible adiabatic (e.g., for calculating final temperature or if the process was irreversible), but they are not required for this specific calculation.

Explanation of the solution:

For a reversible adiabatic process:

1. Adiabatic implies no heat exchange ($Q=0$).
2. Reversible implies no entropy generation, so $\Delta S_{univ}=0$.
3. $\Delta S_{sys} = \int \frac{dQ_{rev}}{T}$. Since $dQ_{rev}=0$ for an adiabatic process, $\Delta S_{sys}=0$.
4. $\Delta S_{surr} = \frac{Q_{surr}}{T_{surr}}$. Since $Q_{surr}=0$ for an adiabatic process, $\Delta S_{surr}=0$.
5. $\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} = 0 + 0 = 0$.