# How do I determine entropy change between two states if given the two temperatures and the two pressures?

## For instance when one mole of a monoatomic ideal gas is taken around a Carnot cycle. At point A, the pressure is P1 and the temperature is 500 K. At point B, the pressure is P2 (P1=2*P2) and the temperature is still 500 K. The only information given about point C in the cycle is that it has a temperature of 400 K.

May 14, 2016

CONDITIONS OF THE PROCESS

Note that from point $A$ to point $B$, the temperature is constant, and the pressure halved. If the pressure changes and the temperature doesn't, then according to the ideal gas law, the volume changes as well.

DETERMINING OTHER KNOWN QUANTITIES

Next, consider the ideal gas law for $\text{1 mol}$ of monatomic ideal gas ($n = 1$):

$\setminus m a t h b f \left(P V = n R T\right)$

$\to P V = R T$

$\to$ ${P}_{2} {V}_{2} = R T$
$\to$ ${P}_{1} {V}_{1} = 2 {P}_{2} {V}_{1} = R T$

We see that these are equal:

$2 \cancel{{P}_{2}} {V}_{1} = \cancel{{P}_{2}} {V}_{2}$

$\implies \textcolor{g r e e n}{{V}_{2} = 2 {V}_{1}} ,$ so the volume doubled.

FINDING SOMETHING WE DO KNOW

We know how the volume changed. Let's focus on that right now. Recall that the change in volume relates back to the expression for reversible (efficient) work:

$\setminus m a t h b f \left({w}_{\text{rev}} = - {\int}_{{V}_{1}}^{{V}_{2}} P \mathrm{dV}\right)$

Now, since the pressure is NOT constant, let's substitute for $P$.

$\textcolor{b l u e}{{w}_{\text{rev}}} = - {\int}_{{V}_{1}}^{{V}_{2}} \frac{R T}{V} \mathrm{dV}$

$= - R T {\int}_{{V}_{1}}^{{V}_{2}} \frac{1}{V} \mathrm{dV}$

$= - R T \left(\ln | {V}_{2} | - \ln | {V}_{1} |\right)$

$= \textcolor{b l u e}{- R T \ln | {V}_{2} / {V}_{1} |}$

At this point, we can find the work. Great. So how does it relate back to the entropy?

RELATING WORK BACK TO ENTROPY

Consider the first law of thermodynamics:

$\setminus m a t h b f \left(\Delta U = {q}_{\text{rev" + w_"rev}}\right)$

For an isothermal process (like here!), $\Delta T = 0$.

Any monatomic ideal gas would have $\Delta U$, the internal energy, equal to $\setminus m a t h b f \left(0\right)$ with this consideration because the internal energy for an ideal gas is solely dependent on the temperature.

Thus:

$\textcolor{g r e e n}{{q}_{\text{rev" = -w_"rev}}}$

Next, you'll have to recall that one of the definitions of entropy is:

$\setminus m a t h b f \left(\Delta S = \frac{{q}_{\text{rev}}}{T}\right)$

where $S$ is entropy, ${q}_{\text{rev}}$ is reversible (efficient) heat flow, and $T$ is the current temperature. That means...

$\Delta S = \frac{- {w}_{\text{rev}}}{T}$

$= \frac{R \cancel{T} \ln | {V}_{2} / {V}_{1} |}{\cancel{T}}$

$= R \ln | {V}_{2} / {V}_{1} |$

Now, we said that the volume of the gas doubled because the pressure of the gas halved at a constant temperature, so naturally we expect the entropy to be positive (as it should be).

What we get for $\setminus m a t h b f \left(\text{1 mol}\right)$ of monatomic ideal gas is:

$\textcolor{b l u e}{\Delta S = R \ln 2 > 0}$