Following problem is based on thermodynamic law. Answer the questions given at the end of it. The second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transition and chemical equilibrium.
Three moles of `CO_(2)` gas expands isothermally (in thermal contact with the surroundings, temperature `=15^(@)` C) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are 10.0 L and 30.0 L, respectively.
`Deltas_("surroundings")` is

Options

A. `27.4 JK^(-1)`
B. `+6.94 JK^(-1)`
C. `0.00 JK^(-1)`
D. `-6.94 JK^(-1)`

Answer

To solve the problem, we need to calculate the change in entropy of the surroundings (ΔS_surroundings) when three moles of CO₂ gas expands isothermally from an initial volume of 10.0 L to a final volume of 30.0 L at a constant temperature of 15°C (which is 288.15 K). The external pressure is given as 1.00 bar.

### Step-by-step Solution:

1. **Convert the temperature to Kelvin:**
\[ T = 15°C + 273.15 = 288.15 \, K \]

**Hint:** Always convert Celsius to Kelvin by adding 273.15.

2. **Calculate the work done (W) during the isothermal expansion:**
The work done by the gas during isothermal expansion against a constant external pressure can be calculated using the formula:
\[ W = -P_{ext} \Delta V \]
where \( \Delta V = V_f - V_i = 30.0 \, L - 10.0 \, L = 20.0 \, L \).

Convert the volume from liters to cubic meters (1 L = 0.001 m³):
\[ \Delta V = 20.0 \, L \times 0.001 \, m³/L = 0.020 \, m³ \]
The external pressure \( P_{ext} = 1.00 \, bar = 100,000 \, Pa \).

Now calculate the work done:
\[ W = -100,000 \, Pa \times 0.020 \, m³ = -2000 \, J \]

**Hint:** Remember that work done by the system is negative when the system expands.

3. **Calculate the heat absorbed by the system (Q):**
For an isothermal process, the heat absorbed by the system is equal to the work done on the surroundings:
\[ Q = -W = 2000 \, J \]

**Hint:** In an isothermal process for an ideal gas, the heat absorbed by the gas is equal to the work done by the gas on the surroundings.

4. **Calculate the change in entropy of the surroundings (ΔS_surroundings):**
The change in entropy of the surroundings can be calculated using the formula:
\[ \Delta S_{surroundings} = -\frac{Q}{T} \]
Substituting the values:
\[ \Delta S_{surroundings} = -\frac{2000 \, J}{288.15 \, K} \approx -6.95 \, J/K \]

**Hint:** The change in entropy of the surroundings is negative when the system absorbs heat.

5. **Final Result:**
The change in entropy of the surroundings is approximately:
\[ \Delta S_{surroundings} \approx -6.95 \, J/K \]

### Summary of the Steps:
- Convert temperature to Kelvin.
- Calculate the work done using the change in volume and external pressure.
- Determine the heat absorbed by the system.
- Calculate the change in entropy of the surroundings using the heat and temperature.