Ideal gas and changes in Thermodynamic Properties

Discuss the Expansion of an ideal gas and changes in Thermodynamic Properties.

We can calculate the change in change in thermodynamic properties like q , w , ∆U , ∆H with the help of first law of thermodynamics. The expansion can be isothermal which can be reversible or irreversible or it can be adiabatic which can also be reversible or irreversible. This is explained below:

Isothermal Expansion:
Calculation of ∆U:
During the expansion of isothermal process, temperature remains constant throughout the process. As internal energy U for an ideal gas depends only on the temperature. As process is isothermal, hence temperature is constant. This means internal energy U remains constant of the gas.

That is ∆U = 0

Calculation of ∆H:
As we know that H = U + PV
Therefore ∆H = ∆ (U + PV) = ∆U + ∆PV
= ∆U + ∆nRT
Since, for an isothermal process, temperature is constant as well as internal energy U remains constant of the gas. That is ∆T as well as ∆U are equal to zero,
Hence,
∆H = 0

Calculation of q and w:
From first law of thermodynamics, ∆U = q + w
As for isothermal process, ∆U = 0
Hence,
W= -q

This means in an isothermal expansion, the work is done at the expense of the heat absorbed.

The magnitude of w and q depends in which the expansion process is carried out, i.e., whether it is carried reversibly or irreversibly.