One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (∆Ssurr) in JK1is (1 L atm = 101.3 J). Why cant we use ∆S = nrln(v2/v1) directly. Why do we have to calculate q using first law?

To tackle this question, we need to understand the relationship between entropy, heat transfer, and the first law of thermodynamics. In this scenario, we have an ideal gas expanding isothermally, and while we might be tempted to use the formula for entropy change directly, there are important considerations regarding the heat exchange with the surroundings that we must account for.

The Concept of Entropy Change

Entropy (S) is a measure of the disorder or randomness in a system. For an ideal gas undergoing an isothermal expansion, the change in entropy of the gas can be expressed as:

• ∆S = nR ln(V2/V1)

Here, n is the number of moles, R is the ideal gas constant, and V1 and V2 are the initial and final volumes, respectively. This formula assumes that the process is reversible and that the system is isolated from its surroundings.

Why We Can't Use ∆S Directly

In your scenario, the gas is expanding against a constant external pressure of 3.0 atm. This means that the gas is doing work on the surroundings, and we need to consider how much heat is exchanged between the gas and its surroundings during this process. The first law of thermodynamics states:

• ΔU = q - W

Where ΔU is the change in internal energy, q is the heat added to the system, and W is the work done by the system. In an isothermal process for an ideal gas, the internal energy change (ΔU) is zero because the temperature remains constant. Therefore, we can simplify the equation to:

• 0 = q - W

This leads us to conclude that:

• q = W

Calculating Work Done

The work done by the gas during the expansion can be calculated using the formula:

• W = P_ext (V2 - V1)

Substituting the given values:

• P_ext = 3.0 atm
• V1 = 1.0 L
• V2 = 2.0 L

First, we convert the pressure from atm to J/L using the conversion factor (1 L atm = 101.3 J):

• P_ext = 3.0 atm × 101.3 J/L = 303.9 J/L

Now, we can calculate the work done:

• W = 303.9 J/L × (2.0 L - 1.0 L) = 303.9 J

Finding Heat Transfer (q)

Since we established that q = W, we have:

• q = 303.9 J

Change in Entropy of the Surroundings

The change in entropy of the surroundings (∆Ssurr) can be calculated using the formula:

• ∆Ssurr = -q/T

Substituting the values we have:

• ∆Ssurr = -303.9 J / 300 K = -1.013 J/K

Final Thoughts

In summary, we cannot directly use the formula ∆S = nR ln(V2/V1) because it does not account for the heat exchange with the surroundings during the expansion against a constant pressure. Instead, we must calculate the work done and use the first law of thermodynamics to find the heat transfer, which allows us to accurately determine the change in entropy of the surroundings. This approach ensures that we consider the entire thermodynamic system and its interactions with the environment.