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

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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.

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.

##### 1 Answer

**CONDITIONS OF THE PROCESS**

Note that from point **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

#\mathbf(PV = nRT)#

#-> PV = RT#

#-># #P_2V_2 = RT#

#-># #P_1V_1 = 2P_2V_1 = RT#

We see that these are equal:

#2cancel(P_2)V_1 = cancel(P_2)V_2#

**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**:

#\mathbf(w_"rev" = -int_(V_1)^(V_2) PdV)#

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

#color(blue)(w_"rev") = -int_(V_1)^(V_2) (RT)/VdV#

# = -RTint_(V_1)^(V_2) 1/VdV#

#= -RT(ln|V_2| - ln|V_1|)#

#= color(blue)(-RT 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**:

#\mathbf(DeltaU = q_"rev" + w_"rev")#

For an **isothermal** process (like here!),

Any monatomic ideal gas would have **the internal energy for an ideal gas is solely dependent on the temperature**.

Thus:

#color(green)(q_"rev" = -w_"rev")#

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

#\mathbf(DeltaS = (q_"rev")/T)#

where *reversible* (efficient) **heat flow**, and **temperature**. That means...

#DeltaS = (-w_"rev")/T#

#= (Rcancel(T) ln|V_2/V_1|)/cancel(T)#

#= Rln|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 **of monatomic ideal gas** is:

#color(blue)(DeltaS = Rln2 > 0)#