Picturing Entropy: Some Examples

The first law of thermodynamics considers interpretations of the law of conservation of energy and how it applies to the system (the reaction under consideration). The second law, however, is interested in the measurement of randomness in a system, the entropy. Before we get down to the nitty-gritty on what that actually means, let's examine some ways in which entropy influences the world around us.

Example 1: Cup of Tea

Say you add an ice cube to your cup of tea. Aside from the wonderful crackling noise the ice cube makes as it melting, what is really going on? First of the enthalpy of the reaction is increasing as the ice cube requires heat to melt. And then, the entropy is also increasing.

Why? Well, in a solid the molecules have less degrees of rotation, as they solely vibrate in a dense structure, but when they melt to become liquid water, the molecules can move around and slide past one another, which is why if you tip over your tea it will spill all over the floor and will make the mess that it does. These further degrees of freedom increase the amount of randomness in the system. And hence, liquids have greater entropy than solids.

Now, let's consider the steam coming off of your tea cup. As you may have anticipated, gases of a substance have even higher entropy than liquids, because the molecules have greater degrees of freedom than liquids.

So in summary: $$entropy_{solid} < entropy_{liquid} < < entropy_{gas}$$
More degrees of freedom = more entropy

Example 2: Denaturing DNA

 Denaturing DNA = breaking DNA from its classic double helix shape into two separate strands.

 Often in the laboratory denaturation of DNA is done in a test tube by increasing the temperature so that the hydrogen bonds that connect the bases (A,T,G,C) break. This is an endothermic process as it requires input of energy, high temperature and results in something with more entropy (2 strands instead of 1).

Denaturing DNA is also a reversible process, so if you decrease the temperature, the strands will re-anneal to the double-stranded helix structure. This can be investigated more clearly by checking out the Gibb's Free Energy equation $G=H-TS$, which I will discuss in a later post.

 Basically, the take home message is that for any reaction, there is a dialogue between the favourable enthalpy and favourable entropy conditions, which will determine whether the reaction is spontaneous or not.

In example 1 with the tea cup, the ice melting and the steam rising, was all spontaneous at room temperature. However, in example 2, this is not as clear. The reaction is temperature dependent, as temperature is the factor that mediates the relationship between the enthalpy and entropy of the reaction.

 But now you are probably wondering: what are the favourable enthalpy and favourable entropy conditions for a reaction to occur spontaneously? Fair enough.

 For a reaction to occur spontaneously, the it prefers to have negative enthalpy, which is an exothermic process (releases heat into the environment).

 "Okay," you say, "But what if I have a giant stick of dynamite in my hands? That releases a lot of heat because it explodes. So why isn't this reaction spontaneous? Clearly it is an exothermic reaction."

Good point. Recall a potential energy diagram for an exothermic process. Some reactions are required to overcome their activation energy for the reactants to turn into the products. Here is a quick 30 second video to trigger your memory if you have forgotten this concept. Activation Energy Definition So in the case of dynamite, it will not spontaneously combust because it requires a certain input of energy - the spark to ignite it - to blow up and overcome its activation energy.

 So, we've covered the spontaneous conditions for enthalpy, but what about entropy? The system is favours states of greater disorder, so the entropy increases positively. A way representing this formally is defining our system (because we can take a system to be as small or as big as we like, as we are the ones who define it) as the universe. Since the universe is an isolated system, we know that $$\Delta{S}_{universe}\geq{0}$$

 Now, this explains why certain reactions with respect to a smaller system can occur at all. As long as the $\Delta{S}_{universe}$ remains positive, which it will because of the massive size of the system, a small negative blip where $\Delta{S}_{system}\leq{0}$, will simply not matter as it is insignificant in the grand scheme of things. We represent this by $$\Delta{S}_{universe}+\Delta{S}_{system}\geq{0}$$ or $$\Delta{S}_{system}+\Delta{S}_{surroundings}\geq{0}$$

 Also, note that a reversible process is defined by $$\Delta{S}_{system}=-\Delta{S}_{surroundings}$$

 All right, this leads us to example 3.

 Example 3: An Endothermic Reaction

 $$\text{N}_2\text{O}_4\longrightarrow 2\text{NO}_2$$

 So what is happening here? Dinitrogen dioxide on the left hand side of the equation is reacting to become the product nitrogen monoxide. The key thing to notice is that there is only 1 mole of reactant to 2 moles of products, so even though this reaction absorbs energy and is endothermic, and hence should be unfavourable, the positive increase in entropy (more products than reactants, therefore more disorder) favours the reaction to proceed. In this case, the drive of the entropy overcomes the enthalpy of the reaction.

In the next post I will discuss a statistical representation of entropy, so stay tuned! :)

An Introduction to Entropy

What is entropy? A measure of disorder can seem abstract and somewhat difficult to envision at times. Can I really blame the state of my apartment with unwashed dishes and dirty laundry strewn on the floor on its agency - or does this have to do more with the subjective nature of personality traits? Can one really measure quantitively disorder when colloquially we refer to it in quantitative terms?

I am very fond of the topic of entropy. In high school, I used to tell horror stories about the eventual heat death of the universe ad naseum. Eventually all of the particles would exist in a solid, perfect crystal state, where no energy could rescue them from their eternal plight.

My musician father was fascinated by entropy while listening to CBC radio program Ideas. Many a skype conversation, he will pounce the topic upon me, and question continuously its definition and properties. Once we devised a system of cells in the human body, existing under ideal conditions (eg: no negative effects of the external environment such as a blow to some tissue, or disease), with only the natural aging process as a factor. Applying the statistical explanation of entropy, we were able to determine it was statically unlikely for all of the cells in an common region of the body to spontaneously die, when otherwise undisturbed.

Walking one evening with a someone special, in the crisp fall Montréal air, we discussed how if you have an empty chamber attached to a flask filled with particles of gas, and opened the flask to the empty chamber, the gas particles would distribute themselves so they filled every state in the chamber. The movement of our postulated system of order to one of disorder mirrored that of the Big Bang. We found it disturbing that the transition from order to disorder was mediated by an external event of the person opening the stopcock. Does this part of the analogy translate over to the scenario of the Big Bang as well or not? Where did the initial stimulus for the start of the universe come from?

Entropy can lead to interesting conversations with people of all different backgrounds, and also contribute to an enhanced understanding of the behaviour of the universe. For these reasons, I am going to develop the concept of entropy over a series of posts, discussing both statistical and thermodynamic interpretations of entropy, and employing both qualitative and quantitative means throughout my explanations.

These posts will be dedicated to my dad because I have always promised to explain entropy throughly to him, but we never have sat down properly to discuss it.

Ideal Gases: Reversible Adiabatic Process

For a reversible adiabatic process, the system is moving between two different isotherms, so the temperature is changing. Also, the both volume and pressure are changing. So $P_1V_1T_1 \rightarrow P_2V_2T_2$.

The work done in an adiabatic process is less than that of an isothermal process, as an isothermal process is the maximum work possible to be done on the system. $w_{adiabatic} < w_{isothermal}$

For a reversible adiabatic process, there is no heat transfer to the system by definition, thus $q=0$. Since $\Delta{U}=q+w$, then $\Delta{U}=w=-PdV$.

For the following derivation, to make our lives a bit easier, we are going to consider the case where $n=1$. $$dU=-PdV$$ $$C_vdT=-PdV$$ $$C_vdT+PdV=0$$ $$C_vdT+RTdV/V=0$$ $$\text{as } P=RT/V$$ $$\text{divide equation by } T$$ $$C_vdT/T+RT/TdV/V=0/T$$ $$C_vdT/T+RdV/V=0$$ $$C_v\int_{T_1}^{T_2}dT/T+R\int_{V_1}^{V_2}dV/V=0$$ $$C_vln(T_2/T_1)+Rln(V_2/V_1)=0$$ $$\text{recall that }R=C_p-C_v$$ $$\text{divide by }C_v$$ $$C_v/C_vln(T_2/T_1)+(C_p-C_v)/C_vln(V_2/V_1)=0$$ $$ln(T_2/T_1)+(C_p/C_v+1)ln(V_2/V_1)=0$$ $$\text{let } C_p/C_v =\gamma$$ $$ln(T_2/T_1)+ln(V_2/V_1)^{\gamma-1}=0$$ $$ln(T_2/T_1)=ln(V_1/V_2)^{\gamma-1}$$ $$T_2/T_1=(V_1/V_2)^{\gamma-1}$$ $$\text{since } PV=RT$$ $$(P_2V_2)/(P_1V_1)=(V_1/V_2)^{\gamma-1}$$ $$P_2/P_1=(V_1/V_2)^{\gamma-1}(V_1/V_2)$$ $$P_2/P_1=(V_1/V_2)^{\gamma}$$ $$P_1V_1^{\gamma}=P_2V_2^{\gamma} \text{ where } \gamma = C_p/C_v$$ 

To check our work, we can see if this applies to the ideal gas law. $$P_1V_1^{\gamma}=P_2V_2^{\gamma}$$ $$PV^{\gamma}=RTV^{\gamma}/V$$ $$PV^{\gamma}=RTV^{\gamma-1}$$ $$PV^{\gamma}/V^{\gamma}=RTV^{\gamma-1}/V^{\gamma}$$ $$P=RT/V$$ The gamma term can also be related to the number of degrees of freedom,$f$, in a gas. $$\gamma = C_p/C_v = \Delta{H}/\Delta{U}=(f+2)R/fR= (f+2)/f$$

To check out $\Delta{H}$, the entropy, we consider $$\Delta{H}=\Delta{U}+\Delta{PV}=w+q+\Delta{PV}=w+R\Delta{T}$$ $$\Delta{H}=C_v\Delta{T}+R\Delta{T}$$ $$\Delta{H}=(C_v+R)\Delta{T}$$ $$\Delta{H}=C_p\Delta{T}$$

In summary:
$$q=0$$ $$\Delta{U}=w=-PdV$$ $$\Delta{H}=C_p\Delta{T}$$ $$P_1V_1^{\gamma}=P_2V_2^{\gamma} \text{ where } \gamma = C_p/C_v$$

Ideal Gases: Reversible Isothermal Process

Consider system with an ideal frictionless piston to apply and reduce pressure and change the volume. Recall that for an isothermal process that is reversible, the $P_{gas}=P_{ext}$. In this case the work in = heat out. Also, there is no net change in both $\Delta{U}$ and $\Delta{H}$, because the process occurs along the same isotherm, so $\Delta{T}=0$, as $T_1=T_2$.

 Therefore (a bit redundant, but we'll show the equations anyway):
 $$\Delta{U}=C_v(T_2-T_1)=0$$ $$\Delta{H}=C_p(T_2-T_1)=0$$

 Recall, the equation of work of an ideal gas $$w=RTln(V_1/V_2)$$

 Since the heat out is equal to the work in, the heat is given by $$q=-w=-RTln(V_1/V_2)=RTln(V_2/V_1)$$

 Note about the Concentration:

 When the number of moles are not equal to one, we cannot ignore the $n$ term. So when have some $P_1=n_1/V_1RT=C_1RT$ and $P_2=n_2/V_2RT=C_2RT$ where $C$ is the concentration.

Therefore, the ideal gas equation for work can also be expressed by $$w=nRTln(V_1/V_2)=NRTln(P_2/P_1)\text{ as } P_2/P_1=V_1/V_2$$ and $$w=nRTln(C_2/C_1)$$

 These equations come directly from the definitions of the terms. If they seem a bit confusing, refer to earlier physical chemistry posts on this blog, which can be found via the table of contents. Or if not, check wikipedia or something. :)

Ideal Gases: Isochoric Process

An isochoric process occurs where the volume is constant, $V_1=V_2$. This means that for the ideal gas law, $PV=nRT$, $T \propto P$.

 Since $\Delta{V}=0$, then the relationship $w=-P\Delta{V}=0$ as well.

Examining the heat at constant volume, we find
$q_v=C_v\int_{T_1}^{T_2}dT=C_v(T_2-T_1)$

 Now, when we look at the internal energy
 $\Delta{U}=q_v+w=q_v=C_v(T_2-T_1)$ 

And finally, the entropy for when $n=1$, with
 $$\Delta{H}=\Delta{U}+\Delta(PV)$$ $$\Delta{H}=q_v+\Delta{RT}$$ $$\Delta{H}=C_v(T_2-T_1)+R(T_2-T_1)$$ $$\Delta{H}=(C_v+R)(T_2-T_1)$$

So, in summary, for an isochoric process $$w_{rev}=0$$ $$q_v=\delta{U}=C_v(T_2-T_1)$$ $$\Delta{H}=(C_v+R)(T_2-T_1)$$

Note, that like for isobaric processes, isochoric processes occur on two different isotherms, as the temperature is not constant. These two isotherms of the isochoric process can be the same isotherms as for an isobaric process, however the work and heat supplied vary.

Ideal Gases: Isobaric Process

Ideal Gases Isobaric Processes

When examining our favourite ideal gas law of $PV=nRT$, we are going to consider a case where the pressure remains constant. Such a case is referred to as an isobaric process, and $P_1 = P_2$.

For example, you have a balloon, with some initial pressure, volume, and temperature. Say that you leave the balloon outside in the cold Montréal winter overnight. In the morning, you would probably see that the balloon shrank. What happened? Well, the atmospheric pressure remained constant, but the temperature changed, dropping to some infernal degree in the negatives. So since the pressure and moles can be eliminated from the ideal gas equation, we have the relationship $V\propto T$. As the temperature decreased overnight, so must the volume to compensate.

Now, enough of that. Let's investigate further to see what happens to the PV work in an isobaric process:

$$w_{rev} = -\int_{V_1}^{V_2}P_1dV
$$ $$w_{rev}= -P_1(V_2-V_1)=+P_1(V_1-V_2)
$$ $$w_{rev}=P_1(RT_1/P_1-RT_2/P_1)
$$ $$w_{rev}=R(T_1-T_2)
$$

 Note that $V_1>V_2$, and also $n=1$.

 The heat is given by $heat=q_p =\int_{T_1}^{T_2}C_Pd = C_p(T_2-T_1)$

 The change in internal energy is given by 
$$\Delta{U}=q_p+w$$ $$\Delta{U}=C_p(T_2-T_1)-R(T_2-T_1)$$ $$\Delta{U}=(C_p-R)(T_2-T_1)$$ $$\Delta{U}=C_v(T_2-T_1)$$

 In summary for an isobaric process, $$q_p=C_p(T_2-T_1)$$ $$w_{rev}=-R(T_2-T_1)$$ $$\Delta{H} = q_p = C_p(T_2-T_1)$$ $$\Delta{U}=C_v(T_2-T_1)$$

 This process occurs on two different isotherms.