Understanding Entropy Change in an Isothermal Process

Entropy is a measure of the disorder or randomness in a system. The change in entropy (ΔS) quantifies how this disorder changes during a process. For a reversible process, the change in entropy of a system is given by the heat transferred (ΔQ) divided by the absolute temperature (T) at which the transfer occurs.

An isothermal process is one that occurs at a constant temperature, meaning ΔT = 0. For an ideal gas undergoing an isothermal process, the internal energy change (ΔU) is zero because internal energy depends only on temperature for an ideal gas.

According to the First Law of Thermodynamics, ΔQ = ΔU + ΔW, where ΔW is the work done by the system. Since ΔU = 0 in an isothermal process for an ideal gas, we have ΔQ = ΔW.

For a reversible process, the infinitesimal heat transfer dQ is equal to the infinitesimal work done dW, which is given by P dV for expansion or compression. So, dQ = P dV.

The change in entropy is given by the integral of $\frac{dQ}{T}$. For an isothermal process, T is constant, so:

$\Delta S = \int \frac{dQ}{T} = \frac{1}{T} \int dQ = \frac{Q}{T}$

For a reversible isothermal process of an ideal gas (assuming n moles), the work done is given by:

$W = \int_{V_1}^{V_2} P dV$

Using the ideal gas law, PV = nRT, we can write P = $\frac{nRT}{V}$. Substituting this into the work equation:

$W = \int_{V_1}^{V_2} \frac{nRT}{V} dV$

Since T is constant in an isothermal process, we can take nRT out of the integral:

$W = nRT \int_{V_1}^{V_2} \frac{1}{V} dV = nRT [\ln V]_{V_1}^{V_2} = nRT (\ln V_2 - \ln V_1)$

$W = nRT \ln\left(\frac{V_2}{V_1}\right)$

Since ΔQ = W for a reversible isothermal process, the heat transferred is $Q = nRT \ln\left(\frac{V_2}{V_1}\right)$.

Now we can find the change in entropy:

$\Delta S = \frac{Q}{T} = \frac{nRT \ln\left(\frac{V_2}{V_1}\right)}{T} = nR \ln\left(\frac{V_2}{V_1}\right)$

If we consider the change in entropy per mole, or if the formulas in the options are implicitly for n=1 mole, then the change in entropy is:

$\Delta S = R \ln\left(\frac{V_2}{V_1}\right)$

We can also express this in terms of pressure. For an isothermal process of an ideal gas, Boyle's Law applies: $P_1V_1 = P_2V_2$. This means $\frac{V_2}{V_1} = \frac{P_1}{P_2}$.

Substituting this ratio into the entropy change equation:

$\Delta S = R \ln\left(\frac{P_1}{P_2}\right)$

Using the property of logarithms that $\ln\left(\frac{a}{b}\right) = -\ln\left(\frac{b}{a}\right)$, we can rewrite this as:

$\Delta S = -R \ln\left(\frac{P_2}{P_1}\right)$

Now let's compare this result with the given options:

• Option 1: $-R \ln\left(\frac{V_2}{V_1}\right)$ - This is incorrect.
• Option 2: $R \ln\left(\frac{P_2}{P_1}\right)$ - This is incorrect, missing the negative sign.
• Option 3: $-R \ln\left(\frac{P_2}{P_1}\right)$ - This matches our derived formula.
• Option 4: $-R \ln\left(\frac{V_2}{V_1}\right) + R \ln\left(\frac{P_2}{P_1}\right)$ - This expression does not represent the total entropy change.

Therefore, the change in entropy in an isothermal process for an ideal gas is given by $-R \ln\left(\frac{P_2}{P_1}\right)$ (assuming 1 mole or the formula is per mole).