To solve this problem, let's proceed step-by-step, starting with the necessary information and using the ideal gas law to determine the number of moles. Then, we'll calculate the entropy change for both the expansion and heating processes.

Given Data:

1. Initial Temperature, $T_1$ = 25°C (convert to Kelvin: $T_1 = 25 + 273.15 = 298.15 \, \text{K}$)
2. Initial Pressure, $P_1 = 0.101 \, \text{MPa} = 101 \, \text{kPa}$
3. Initial Volume, $V_1 = 0.5 \, \text{L}$
4. Final Temperature, $T_2 = 100°C$ (in Kelvin: $T_2 = 100 + 273.15 = 373.15 \, \text{K}$)
5. Final Volume, $V_2 = 1.0 \, \text{L}$

Step 1: Calculate the Number of Moles ( $n$ )

Using the ideal gas law: $PV = nRT$ where:

• $P$ is the pressure,
• $V$ is the volume,
• $n$ is the number of moles,
• $R$ is the ideal gas constant ( $R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}$ ), and
• $T$ is the temperature.

Plugging in the initial conditions: $n = \frac{P_1 V_1}{R T_1} = \frac{(101 \times 10^3 \, \text{Pa}) \cdot (0.5 \times 10^{-3} \, \text{m}^3)}{8.314 \cdot 298.15}$ Calculating this gives: $n \approx 0.0204 \, \text{moles}$

Step 2: Determine the Total Entropy Change ($\Delta S$)

The entropy change ($\Delta S$) for the process of expansion and heating can be divided into two parts:

1. Entropy change due to heating from $T_1$ to $T_2$ at constant volume.
2. Entropy change due to isothermal expansion from $V_1$ to $V_2$ at the higher temperature $T_2$.

2.1: Entropy Change due to Heating at Constant Volume

The formula for entropy change when heating at constant volume is: $\Delta S_{\text{heating}} = n C_V \ln \frac{T_2}{T_1}$ For a monatomic ideal gas (such as Argon), the molar heat capacity at constant volume, $C_V = \frac{3}{2} R$.

Substituting values: $\Delta S_{\text{heating}} = (0.0204) \cdot \left( \frac{3}{2} \cdot 8.314 \right) \ln \frac{373.15}{298.15}$ Calculating this gives: $\Delta S_{\text{heating}} \approx 0.209 \, \text{J/K}$

2.2: Entropy Change due to Isothermal Expansion

The formula for entropy change due to isothermal expansion at temperature $T_2$ is: $\Delta S_{\text{expansion}} = n R \ln \frac{V_2}{V_1}$ Substituting the values: $\Delta S_{\text{expansion}} = (0.0204) \cdot 8.314 \ln \frac{1.0}{0.5}$ Calculating this gives: $\Delta S_{\text{expansion}} \approx 0.1176 \, \text{J/K}$

Step 3: Total Entropy Change

The total entropy change for the process is the sum of the entropy changes for heating and expansion: $\Delta S_{\text{total}} = \Delta S_{\text{heating}} + \Delta S_{\text{expansion}}$ $\Delta S_{\text{total}} = 0.209 + 0.1176 = 0.3266 \, \text{J/K}$

Answer

The total entropy change $\Delta S_{\text{total}}$ for the process is approximately: $\Delta S_{\text{total}} \approx 0.327 \, \text{J/K}$

Would you like further clarification on any part of the calculation?

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Related Questions

1. How would the entropy change differ if the gas was a diatomic rather than a monatomic gas?
2. How does the ideal gas law apply to the entropy change in this scenario?
3. What if the volume expanded to 2 liters instead of 1 liter?
4. How does the path taken by the gas affect the entropy calculation?
5. How would the calculation change if the process was adiabatic instead of isothermal?

Tip

When calculating entropy changes, consider breaking down the process into simpler steps (e.g., isothermal and isochoric steps) as entropy is a state function and allows for flexibility in choosing intermediate steps.