Please wait for the animation to completely load. There is a time delay—since the system must be in equilibrium—before the initial change of state occurs.

**In this animation N = nR** (i.e., k_{B} = 1). This, then, gives the
ideal gas law as PV = NT. Restart

In thermodynamic processes the entropy depends not on the path taken but on the end points. It is a "state function" (in contrast to heat and work, which depend on the process). Since Q = ΔU + W and ΔU = (3/2)nRΔT (for a monatomic gas) ,

ΔS = ∫dQ/T = ∫(3/2)nRdT/T + ∫PdV/T =
nR[(3/2)ln(T_{f}/T_{i}) + ln(V_{f}/V_{i})].

Thus, ΔS = (3/2)N ln(T_{f}/T_{i}) + N ln(V_{f}/V_{i})
for an ideal monatomic gas (note that ln represents the natural log, base e).

In the animations, note that the area under the PV diagram is equal to the work.

What is the work done in each case?

What is the heat absorbed or released in each case?

What is the area under the associated TS diagram? (Note that the choice of the initial entropy is arbitrary.)

How does the change in entropy compare for the three processes?

Compare your measurements from the graphs to the calculated values found from using the equation above for an ideal monatomic gas.

Another way to measure the change in entropy is to use Q = mcΔT or, for a gas, Q = CNΔT. In this case,

ΔS = CN ln(T_{f}/T_{i}).

Show that, for the isobaric expansion where C = C

_{P }= (5/2), you get this change in entropy.

Exploration authored by Anne J. Cox.

HTML updated for JavaScript by Aidan
Edmondson.

Physlets were developed at Davidson College and converted from Java to
JavaScript using the SwingJS
system developed at St. Olaf College.