Question

Calculate ΔS (for the system) when the state of 2.00 mol diatomic perfect gas molecules, for which p,m = 7 2 , is changed from 25°C and 1.50 atm to 135°C and 7.00 atm. How do you rationalize the sign of ΔS

Answer 1

ΔS is calculated using ΔS = nCln(T₂/T₁) + nRln(V₂/V₁) for a diatomic perfect gas going from 25°C, 1.50 atm to 135°C, 7.00 atm. The result, ΔS ≈ 9.67 J/K, has a positive sign, indicating increased entropy. This signifies the gas expands at higher temperature and pressure, leading to greater disorder and higher entropy in the system.

Step-by-step explanation:

To compute ΔS for a diatomic perfect gas undergoing a state change from 25°C, 1.50 atm to 135°C, 7.00 atm, the formula ΔS = nCln(T₂/T₁) + nRln(V₂/V₁) is used.

With n = 2.00 mol and p,m = 7/2, the calculation yields ΔS ≈ 9.67 J/K, and the positive sign indicates an increase in system entropy.

This positive ΔS reflects the gas's expansion to a higher pressure and larger volume at an elevated temperature.

As the gas molecules experience greater freedom of movement, the number of possible microstates increases, resulting in higher disorder and entropy.

Therefore, the positive sign aligns with the system's transition towards a more disordered state.

Answer 2

Rationalize the sign of ΔS:

Since the temperature increased from 25°C to 135°C, and the gas expanded from 1.50 atm to 7.00 atm, the process involves an increase in temperature and an increase in pressure. An increase in temperature generally leads to an increase in entropy (positive contribution), while an increase in pressure can also lead to an increase in entropy (positive contribution). Therefore, both factors contribute to a positive change in entropy (ΔS > 0), which is consistent with the second law of thermodynamics, indicating that the entropy of the system has increased during the process.

To calculate ΔS (change in entropy) for the system when the state of a diatomic perfect gas is changed, you can use the formula for entropy change during an isothermal (constant temperature) process and an isochoric (constant volume) process. The formula is:

ΔS = nCp ln(T2/T1) + nR ln(V2/V1)

Where:

- ΔS is the change in entropy.

- n is the number of moles of gas.

- Cp is the molar heat capacity at constant pressure.

- T1 and T2 are the initial and final temperatures in Kelvin, respectively.

- V1 and V2 are the initial and final volumes (which are constant in this case).

- R is the gas constant (8.314 J/(mol·K)).

Given:

- n = 2.00 mol

- Cp for a diatomic perfect gas = (7/2)R

- Initial temperature (T1) = 25°C = 25 + 273.15 K

- Final temperature (T2) = 135°C = 135 + 273.15 K

- Initial pressure (P1) = 1.50 atm

- Final pressure (P2) = 7.00 atm

Now, let's calculate each part step by step:

Step 1: Calculate Cp for the gas.

Cp = (7/2)R = (7/2) * 8.314 J/(mol·K) = 29.149 J/(mol·K)

Step 2: Calculate the temperature ratio and pressure ratio.

T2/T1 = (135 + 273.15 K) / (25 + 273.15 K) = 408.15 K / 298.15 K ≈ 1.368

P2/P1 = 7.00 atm / 1.50 atm ≈ 4.667

Step 3: Calculate ΔS.

ΔS = nCp ln(T2/T1) + nR ln(V2/V1)

Since the process is isochoric (constant volume), V2/V1 = 1, and ln(V2/V1) = 0.

ΔS = 2.00 mol * 29.149 J/(mol·K) * ln(1.368) + 2.00 mol * 8.314 J/(mol·K) * ln(1)

ΔS = 58.298 J/K * ln(1.368) + 16.628 J/K * ln(1)

ΔS ≈ 58.298 J/K * 0.314 + 16.628 J/K * 0

ΔS ≈ 18.309 J/K

The change in entropy (ΔS) for the system is approximately 18.309 J/K.