An ideal gas is compressed at constant temperature. What are the signs ( + , - , or 0) of ∆H, ∆S, and ∆G for the process? Explain.

Verified step by step guidance

1. Identify that the process involves the compression of an ideal gas at constant temperature.
2. Recall that for an ideal gas, the internal energy (U) depends only on temperature. Since the temperature is constant, there is no change in internal energy (∆U = 0).
3. Understand that enthalpy (H) is related to internal energy and pressure-volume work. Since ∆U = 0 and the process is isothermal, ∆H = 0.
4. Recognize that entropy (S) is a measure of disorder. Compression of a gas decreases its volume, leading to a decrease in entropy (∆S < 0).
5. Use the Gibbs free energy equation, ∆G = ∆H - T∆S. Since ∆H = 0 and ∆S < 0, ∆G will be positive (∆G > 0) because the term -T∆S will be positive.

Key Concepts

Enthalpy (∆H)

Enthalpy is a measure of the total heat content of a system. In an isothermal process, such as the compression of an ideal gas at constant temperature, the internal energy remains constant, and any work done on the gas is balanced by heat exchange with the surroundings. For this process, ∆H is 0, indicating no net heat transfer into or out of the system.

Entropy (∆S)

Entropy is a measure of the disorder or randomness in a system. When an ideal gas is compressed, the number of available microstates decreases, leading to a reduction in disorder. Therefore, for this isothermal compression, ∆S is negative, reflecting a decrease in the system's entropy as the gas molecules are forced closer together.

Gibbs Free Energy (∆G)

Gibbs Free Energy combines enthalpy and entropy to determine the spontaneity of a process. The relationship is given by the equation ∆G = ∆H - T∆S. Since ∆H is 0 and ∆S is negative, ∆G will be positive, indicating that the process is non-spontaneous under constant temperature conditions. This means that work must be done on the system to achieve the compression.