Question

Some ideal gas is constrained in the V"part of anadiabatic container of volume V" V$. The rest of the container is vacuum. When the wall between the V"and V$part of the container is removed, the ideal gas will expand and occupy the whole V" V$. The temperature of the gas before the removal of the wall is T. What is the change of entropyof this process

Answer 1

Answer:

Using the log combination rules to reduce the famous Sakur-Tetrode equation, The change in entropy is given as:

∆S = NK*ln(V"V$/V").

Where V"V$ is final Volume (Vf) after constraint's removal,

V" is Initial Volume (Vi) before constraint's removal.

Temperature (T) is constant, Internal Energy, U is constant, N and K have their usual notations

Explanation:

Given in the question, the container is an adiabatic container.

For an adiabatic contain, it does not permit heat to the environment due to its stiff walls. This implies that the Internal Energy, U is kept constant(Q = U). The temperature is also constant (Isothermal). Thus, the famous Sakur-Tetrode equation will reduce to ∆S = NK* In(Vf/Vi).

Vf is the volume after the constraint is removed(Vf = V"V$). Vi is the volume occupied before the constraint is removed (Vi = V")