Question

A unit mass of an ideal gas at temperature \(T\) undergoes a reversible isothermal process from pressure \(P_{1}\) to pressure \(P_{2}\) while losing heat to the surroundings at temperature \(T\) in the amount of \(q .\) If the gas constant of the gas is \(R,\) the entropy change of the gas \(\Delta s\) during this process is \((a) \Delta s=R \ln \left(P_{2} / P_{1}\right)\) \((b) \Delta s=R \ln \left(P_{2} / P_{1}\right)-q / T\) \((c) \Delta s=R \ln \left(P_{1} / P_{2}\right)\) \((d) \Delta s=R \ln \left(P_{1} / P_{2}\right)-q / T\) \((e) \Delta s=0\)

Short answer

Answer: (c) Δs = R ln(P1/P2)

Step-by-step solution

Recall the equation for entropy change for a reversible isothermal process

For a reversible isothermal process, the entropy change of an ideal gas is given by: Δs = nR ln(V2/V1) where n is the number of moles, R is the gas constant, and V1 and V2 are the initial and final volumes, respectively.

Find a relation between volume and pressure for an ideal gas

We know that for an ideal gas, the equation of state is given by: PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since the process is isothermal, T remains constant. Therefore, we can relate the initial and final states as: P1V1 = nRT P2V2 = nRT

Derive a relation between V1 and V2 in terms of P1 and P2

We can divide the two equations obtained in Step 2: P1V1/P2V2 = 1 This simplifies to: V2/V1 = P1/P2

Substitute the relation in the entropy change equation

Now replace V2/V1 in the entropy change equation from Step 1 with the derived relation: Δs = nR ln(P1/P2) Since we are only considering unit mass, n = 1, so the equation becomes: Δs = R ln(P1/P2)

Match the calculated expression with provided options

Comparing this expression with the given options, we find that: Δs = R ln(P1/P2) matches with option (c). Therefore, the correct answer is (c) Δs = R ln(P1/P2).

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental relation in thermodynamics that provides a link between the pressure, volume, temperature, and the number of moles of an ideal gas. It is expressed as:
\( PV = nRT \)
Here, \(P\) stands for pressure, \(V\) represents volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) denotes temperature.
For an isothermal process, the temperature \(T\) remains constant, implying that the product of the pressure and volume at one state will be equal to the product at any other state. This principle is crucial when considering entropy changes in reversible isothermal processes, as it provides the necessary relationship to connect the change in volume to the change in pressure.

Isothermal Process

An isothermal process is a thermodynamic process in which the temperature of the system remains constant throughout. This implies that any heat added or removed from the system is done so in a manner that does not change the system's temperature.
For an ideal gas undergoing an isothermal process, using the ideal gas law, changes in pressure will inversely affect the volume to maintain the same temperature. For instance, if the pressure of the gas increases, the volume must decrease to keep the product \(PV\) constant if the temperature is constant.

Importance in Calculating Entropy Change

In thermodynamics, calculating the change in entropy during an isothermal process requires consideration of this inverse relationship between pressure and volume. As the ideal gas expands or compresses isothermally, it does work on its surroundings or has work done on it, which can be directly related to its entropy change.

Thermodynamic Entropy

Thermodynamic entropy is a measure of the number of specific ways a thermodynamic system can be arranged, commonly understood as a measure of disorder or randomness. In the context of an ideal gas, entropy can be understood as the spread of gas particles.
During a reversible isothermal process, the change in entropy \(\Delta s\) can be calculated using the relationship to volume change as per the ideal gas law, given by:
\(\Delta s = R \ln(\frac{V2}{V1})\)
Since entropy is a state function, it only depends on the initial and final states of the system, not on the path taken.

Connecting Heat Transfer and Entropy

When an ideal gas undergoes an isothermal process, any heat \(q\) transferred to or from the surroundings at temperature \(T\) is directly related to the change in the system's entropy, as heat transfer is one of the mechanisms by which entropy changes. This relationship is often used to determine the entropy change in thermodynamic processes.