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# Thermodynamics article

## Law of Thermodynamics

Thermodynamics is a very important branch of both physics and chemistry. It deals with the study of energy, the conversion of energy between different forms and the ability of energy to do work. As you go through this article, I am pretty sure that you will begin to appreciate the importance of thermodynamics and you will start noticing how laws of thermodynamics operate in your daily lives! There are essentially four laws of thermodynamics.

## Zeroth Law of Thermodynamics

“If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with one another”
Let’s first define what ‘thermal equilibrium’ is. When two systems are in contact with each other and no energy flow takes place between them, then the two systems are said to be in thermal equilibrium with each other. In simple words, thermal equilibrium means that the two systems are at the same temperature. Thermal equilibrium is a concept that is so integral to our daily lives. For instance, let's say you have a bowl of hot soup and you put it in the freezer. What will happen to the soup? The soup will, of course, start cooling down with time. You all know that. And you also probably know that the soup will continue to cool down until it reaches the same temperature as the freezer. Even if you are familiar with this concept, what you may not realize is that this is an excellent example of thermal equilibrium. Here, heat flows from the system at a higher temperature (bowl of soup) to the system at a lower temperature (freezer). Heat flow stops when the two systems reach the same temperature. In other words, now the two systems are in thermal equilibrium with each other and there is no more heat flow taking place between the two systems. Let’s assume we have three systems - system 1, system 2 and system 3. Their temperatures are Tstart subscript, 1, end subscript, Tstart subscript, 2, end subscript and Tstart subscript, 3, end subscript respectively.
The Zeroth law states that if the temperature of system 1 is equal to temperature of system 3, and the temperature of system 2 is equal to temperature of system 3; then the temperature of system 1 should be equal to the temperature of system 2. The three systems are said to be in thermal equilibrium with each other.
That is; if Tstart subscript, 1, end subscript = Tstart subscript, 3, end subscript and Tstart subscript, 2, end subscript = Tstart subscript, 3, end subscript, then Tstart subscript, 1, end subscript = Tstart subscript, 2, end subscript
The zeroth law is analogous to the basic rule in algebra, if A=C and B=C, then A=B.
This law points to a very important fact - ‘temperature affects the direction of heat flow between systems.’ Heat always flows from high temperature to low temperature. Heat flow is mathematically denoted as ‘Q’.

## First Law of Thermodynamics

This law is essentially the ‘law of conservation of energy’. Energy can neither be created nor destroyed; it can just be converted from one form to another.
In simple words, the first law of thermodynamics states that whenever heat energy is added to a system from outside, some of that energy stays in the system and the rest gets consumed in the form of work. Energy that stays in the system increases the internal energy of the system. This internal energy of the system can be manifested in various different forms – kinetic energy of molecules, potential energy of molecules or heat energy (that simply raises the temperature of the system).

### What is internal energy?

It is defined as the sum total of kinetic energy, which comes from motion of the molecules, and potential energy which comes from the chemical bonds that exist between the atoms and any other intermolecular forces that may be present.
First law of thermodynamics is thus conventionally stated as: “The change in internal energy of a closed system is equal to the energy added to it in the form of heat (Q) plus the work (W) done on the system by the surroundings.”
Mathematically, this can be put as
∆Estart subscript, i, n, t, e, r, n, a, l, end subscript = Q + W
Conventional definition of the first law is based on the system gaining heat and the surrounding doing work. The opposite scenario can occur too, in which case the ‘signs’ in the equation will have to be changed appropriately. This will be discussed shortly.
Lots of times you will notice that ∆Estart subscript, i, n, t, e, r, n, a, l, end subscript is also denoted as ∆U.
Now you must be wondering what is meant by a ‘closed system’. Let’s try to understand this through a simple example. Imagine we have two saucepans containing water - 1) with a lid 2) without a lid. Both are kept on a heated stove.
Both the saucepans shown above will absorb heat from the stove and get heated up. So there is exchange of energy taking place in both the cases from the stove (surroundings) to the system (water). But do you notice a difference? Saucepan, with the lid, prevents any addition of mass to it or removal of mass from it; whereas in the case of the saucepan, without the lid, we can easily add coffee and sugar from outside and thus change the mass of the contents. Basically, the lid is preventing matter from entering the saucepan and leaving the saucepan. Here, the saucepan with a lid is an example of a closed system; while the saucepan without a lid is an example of an open system.
Thus, an ‘open system’ can be defined as a system that freely exchanges both energy and matter with its surroundings; while a ‘closed system’ is a system that exchanges only energy with its surroundings, not matter.
PS: Coming back to the sign conventions for the first law of thermodynamics ( ∆Estart subscript, i, n, t, e, r, n, a, l, end subscript = Q + W) ; let’s be ‘very very’ clear of the following norms -
• if heat flows out of the system, then Q will be negative
• if heat flows into the system, then Q will be positive
• if work is done by the system, then W will be negative
• if work is done on the system, then W will be positive
Let’s look at the following example:
We have a gas in a sealed container (closed system; no matter exchange can take place with the surroundings). A piston is attached, on top of which is placed a block of wood. We provide heat (Q) from outside to this system. This heat energy leads to expansion of the gas, which in turn pushes the piston up. So, the gas does work (W) in expanding itself, which results in pushing of the piston.
If you recall from the ideal gas laws; for an ideal gas, work (W) = PV = nRT
Let’s now mathematically define the change in the internal energy of the above system.
Change in internal energy of a system (∆E) = Q + W
Let’s try to apply the rules we talked about earlier,
In this example, *heat flows into the system, so Q will be positive and the work is done by the system (gas in this case), so W will be negative *
Thus, (∆E) = Q – W = Q - P∆V
[Q is the external energy provided, P is the pressure of the gas, and ∆V is the change in volume of the gas]
Now let’s attempt a couple of problems dealing with the first law of thermodynamics. Always be sure to use the correct units while solving any numerical!!!
Problem 1: In an exothermic process, the volume of a gas expanded from 186 mL to 1997 mL against a constant pressure of 745 torr. During the process, 18.6 calories of heat energy were given off. What was the internal energy change for the system in joules? Also, (1 L atm = 101.3 J)
Q = heat given off by system = 18.6 cal = 18.6 x 4.184 J (remember: 1 cal = 4.184 J)
= 77.82 J
Since heat flows out of the system, Q will be negative. So Q = -77.82 J
Work (W) = P∆V, where P = constant pressure of gas, ∆V = change in volume of gas
Let’s first convert pressure and volume into the correct units
760 torr = 1 atm, so 745 torr of pressure = 745/ 760 = 0.98 atm
Volume should be in litres (L), thus ∆V = (1997 – 186) ml = 1811 mL = 1811 X 10start superscript, minus, 3, end superscript L
Thus, W = P∆V = 0.98 atm X 1.811 L = 1.77 atm L
Also, the problem statement gives conversion between atm L and Joules (1 L atm = 101.3 J). So, 1.77 atm L = 1.77 X 101.3 = 179.30 J
In this example, work is being done by the system in expansion of the gas, so W is negative (remember: if work is done by the system, then W will be negative). Therefore, W = -179.30 J So, net change in internal energy of the system (∆U) = Q + W = -77.82 + (-179.30) = -257.12 J
This result indicates that the energy of the gas is decreases by 257.12 J. This means, at the end of the process the gas has less energy than it had in the beginning.
Problem 2: The work done when a gas is compressed in a cylinder is 820 J. At the same time, the system lost 320 J of heat to the surrounding. What is the energy change of this system?
Here the gas is the system. First you must decide the signs of ‘W’ and ‘Q’ using the convention discussed earlier. Work is done on the system, so W = + 820 J and heat is lost by the system, so Q = - 320 J.
Therefore ∆E = Q + W = - 320 J + 820 J = 500 J
This result indicates that the energy of the gas is increased by 500 J. This implies, at the end of the process the gas has more energy than it had in the beginning.

## Second Law of Thermodynamics

This law states “The total change in entropy of a system plus its surroundings will always increase for a spontaneous process”
Entropy is defined as the “measure of disorder or randomness of a system”. Every system wants to achieve a state of maximum disorder or randomness. A commonly observed daily life example of something constantly moving towards a state of randomness is a ‘kid’s room’. Every time mom cleans up the room, within minutes the room looks like this
This is the natural state in which a kid’s room wants to exist ☺
Another commonly encountered entropy driven process is the melting of ice into water. This happens spontaneously as soon as ice is left at room temperature. Ice is a solid with an ordered crystalline structure as compared to water, which is a liquid in which molecules are more disordered and randomly distributed. All natural processes tend to proceed in a direction which leads to a state that has more random distribution of matter and energy.
All of these processes take place spontaneously, meaning that once they start, they will proceed to the end if there is no external intervention. You will never witness the reverse of this process, in which water converts back to ice at room temperature. In other words, it would be inconceivable that this process could be reversed without tampering with the external conditions (you will have to put water in the freezer to force it to form ice). So what determines the direction in which a process will go under a given set of conditions? Now you know the answer - all these processes are driven by entropy.
Mathematically, the second law of thermodynamics can be stated as ∆Sstart subscript, u, n, i, v, e, r, s, e, end subscript = ∆Sstart subscript, s, y, s, t, e, m, end subscript + ∆Sstart subscript, s, u, r, r, o, u, n, d, i, n, g, s, end subscript > 0
Where, ∆Sstart subscript, u, n, i, v, e, r, s, e, end subscript = net change in entropy of the universe
∆Sstart subscript, s, y, s, t, e, m, end subscript = net change in entropy of the system
∆Sstart subscript, s, u, r, r, o, u, n, d, i, n, g, s, end subscript = net change in entropy of the surroundings
Take home message: “Entropy of the universe is constantly increasing
Take home message: “Entropy of the universe is constantly increasing” Entropy is mathematically calculated as Q/ T (heat absorbed or released by the system or surroundings divided by the temperature of the system or surroundings). Entropy is expressed in Joules per Kelvin (J/ K).

### Why is the second law of thermodynamics so important?

Second law of thermodynamics is very important because it talks about entropy and as we have discussed, ‘entropy dictates whether or not a process or a reaction is going to be spontaneous’.
I want you to realize that any natural process happening around you is driven by entropy!!
Let’s take a daily life example:
We drink coffee every day. What happens to our cup of hot coffee in say 10 minutes? The coffee starts getting cold, or in thermodynamic terms you will tell me that the hot coffee gives out heat to the surroundings and in turn the coffee cools down. You are 100% correct, but the thing you might not have realized is that this very obvious daily life phenomenon is governed by “entropy”. Let’s try to prove this mathematically. As shown below, the following two scenarios are possible
The temperature of the surroundings and the coffee are 25 start superscript, o, end superscriptC and 45 start superscript, o, end superscriptC respectively.
Scenario 1: 10 J of heat is absorbed (from the surroundings) by coffee, so surroundings lose 10 J and the system (coffee) gains 10 J. Thus, Qstart subscript, s, y, s, t, e, m, end subscript = + 10 J and Qstart subscript, s, u, r, r, o, u, n, d, i, n, g, s, end subscript = -10 J
∆Sstart subscript, u, n, i, v, e, r, s, e, end subscript = ∆Sstart subscript, s, y, s, t, e, m, end subscript + ∆Sstart subscript, s, u, r, r, o, u, n, d, i, n, g, s, end subscript
= Qstart subscript, s, y, s, t, e, m, end subscript/ Tstart subscript, s, y, s, t, e, m, end subscript + Qstart subscript, s, u, r, r, o, u, n, d, i, n, g, s, end subscript/ Tstart subscript, s, u, r, r, o, u, n, d, i, n, g, s, end subscript
= 10/ (45 + 273) + (-10)/ (25 + 273) [temperatures are to be converted into Kelvin]
= -0.0021 Joules/ Kelvin
This violates the second law of thermodynamics (∆Sstart subscript, u, n, i, v, e, r, s, e, end subscript should be greater than zero), so the above process cannot occur spontaneously.
Scenario 2: 10 J of heat is released (to the surroundings) by hot coffee, so surroundings gain 10 J and system (coffee) loses 10 J. Thus, Qstart subscript, s, y, s, t, e, m, end subscript = - 10 J and Qstart subscript, s, u, r, r, o, u, n, d, i, n, g, s, end subscript = +10 J
∆Sstart subscript, u, n, i, v, e, r, s, e, end subscript = ∆Sstart subscript, s, y, s, t, e, m, end subscript + ∆Sstart subscript, s, u, r, r, o, u, n, d, i, n, g, s, end subscript
= Qstart subscript, s, y, s, t, e, m, end subscript/ Tstart subscript, s, y, s, t, e, m, end subscript + Qstart subscript, s, u, r, r, o, u, n, d, i, n, g, s, end subscript/ Tstart subscript, s, u, r, r, o, u, n, d, i, n, g, s, end subscript
= -10/ (45 + 273) + 10/ (25 + 273) [temperatures have to be converted into Kelvin]
= +0.0021 Joules/ Kelvin
This obeys the second law of thermodynamics (∆Sstart subscript, u, n, i, v, e, r, s, e, end subscript >0), so the above process occurs spontaneously.
In the case of a cup of coffee this was pretty intuitive. But this is not true for chemical reactions. Just by looking at a chemical reaction, we cannot predict if it will take place spontaneously or not. So, calculating the entropy change for that particular reaction becomes important.

### Gibb’s free energy (G): Predictor of spontaneity of a chemical reaction

Gibb’s free energy is defined as ‘the energy associated with a chemical reaction that can be used to do work.’ The free energy (G) of a system is the sum of its enthalpy (H) minus the product of the temperature (T) and the entropy (S) of the system:
G = H - TS
For details on ‘enthalpy’, refer to the article on ‘Endothermic and exothermic reactions’.
Gibbs free energy combines the effect of both enthalpy and entropy. The change in free energy (ΔG) is equal to the sum of the change of enthalpy (∆H) minus the product of the temperature and the change of entropy (∆S) of the system.
∆G = ∆H - T∆S
*ΔG predicts the direction in which a chemical reaction will go under two conditions: (1) constant temperature and (2) constant pressure. *
Rule of thumb: If ΔG is positive, then the reaction is not spontaneous (it requires the input of external energy to occur) and if it is negative, then it is spontaneous (occurs without the input of any external energy).
Let’s attempt a problem involving ΔG
Calculate ΔG for the following reaction at 25 start superscript, o, end superscriptC
NHstart subscript, 3, end subscript(g) + HCl(g) → NHstart subscript, 4, end subscriptCl(s)
but first we need to convert the units for ΔS into kJ/ K (or convert ΔH into J) and temperature into Kelvin. So,
ΔS = −284.8 J/ K = −0.2848 kJ/ K
T = 273.15 K + 25 = 298 K
*(1 kJ = 1000 J)
Now, ΔG = ΔH − TΔS
ΔG = −176.0 kJ − (298 K)(−0. 284.8 kJ/ K)
ΔG = −176.0 kJ − (−84.9 kJ)
ΔG = −91.1 kJ
Answer: Yes, this reaction is spontaneous at room temperature since ΔG is negative.
The beauty of the Gibb’s free energy equation is its ability to determine the relative importance of the enthalpy (ΔH) and the entropy (ΔS) of the system. The change in free energy of the system measures the balance between the two driving forces (ΔH and ΔS), which together determine whether a reaction is spontaneous or not. Let us tabulate this information to make it easier to comprehend ΔG = ΔH − TΔS
Favorable reaction conditionsUnfavorable reaction conditions
∆H < 0∆H > 0
∆S > 0∆S < 0
∆G < 0∆G > 0
• if ∆H < 0 and ∆S > 0, without even doing any calculations you can say that the reaction will be spontaneous because ΔG = ΔH – TΔS will be negative
• if ∆H >0 and ∆S < 0, without even doing any calculations you can say that the reaction will not be spontaneous because ΔG = ΔH – TΔS will be positive
• actual calculations become necessary when out of the two parameters, ∆H and ∆S, one is favorable and the other is not. In such a case, ΔG has to be calculated to predict the spontaneity of the reaction

## Third Law of Thermodynamics

“The entropy of a perfect crystal of any pure substance approaches zero as the temperature approaches absolute zero.”
Whenever I think of this law, I am reminded of the beautiful documentary made on the survival tactics of penguins ‘March of the Penguins’. Those who have watched this documentary will recall: in order to survive the extremely cold weathers of Antarctica (where temperatures approach −80 °F or 210.92 K), penguins form colonies consisting of a group of huddles; and in each colony the penguins are tightly packed and don’t move at all, and all the penguins face in the same direction as shown in the image on the right.
Now assume these penguins to be atoms. Analogous to the penguins, at a temperature of zero Kelvin the atoms in a pure crystalline substance get aligned perfectly and do not move around. Matter is in a state of maximum order (least entropy) when the temperature approaches absolute zero (0 Kelvin). In other words, the entropy of a perfect crystal approaches zero when the temperature of the crystal approaches 0 Kelvin. This is, in fact, the third law of thermodynamics. The consequence of this law is that any thermal motion ceases in a perfect crystal at 0 K. Conversely, if there will be any thermal motion within the crystal at 0 K, it will lead to disorder in the crystal because atoms will start moving around, violating the third law of thermodynamics.

## Want to join the conversation?

• Disorder is the molecular movement of particles in general??
(5 votes)
• The greater the number of microstates of a system the greater the entropy....a microstate specifies all molecular details about the system, including the position and velocity of every molecule. With more available microstates, the entropy of a system increases.
(8 votes)
• For the problem about solving for Delta G where did the ΔS = −284.8 J/ K = −0.2848 kJ/ K come from?

I am unsure where the value is taken from
(4 votes)
• ΔS = −284.8 J/ K = −0.2848 kJ/ K
didnt quite get where this value came from, the innitial J/K thanks to whoever can point this out
(3 votes)
• He looked them up in a table of values; e.g., CRC handbook, NIST website
(2 votes)
• If work is positive when the surroundings do work on the system, then shouldn't heat be positive when it leaves the system into the surroundings. And when heat leaves the surroundings into the system, shouldn't Q be negative?
(2 votes)
• I cannot fully follow what you mean....However, remember that heat leaving the system is always negative....because (if no work is involved) the change in internal energy of the system decreases....If work is positive and q (heat) is negative then the sign of DU (change in internal energy) will depend on the magnitude of work and heat......
(2 votes)
• Doesn't spontaneity depend on enthalpy too (not just entropy) based on the equation for Gibb's free energy? You say that "[Scenario 1] violates the second law of thermodynamics (delta S should be greater than zero), so the above process cannot occur spontaneously." Isn't it correct to say that if the change in entropy is negative, a reaction can still be spontaneous if temperature is low enough and the change in enthalpy is positive?
(2 votes)
• Yes, Sandy it is possible. However, in the definition of spontaneous, it is talking about a system with its surroundings being at a regular temperature, not an extreme case where the temperature is extremely high or low.
(1 vote)
• is entropy related to temperature? if so ,how?
(1 vote)
• Increased temperature = increased entropy
(2 votes)
• On the 9th paragraph, under What is internal energy heading:

"∆E_{internal}
​internal
​​ start subscript, i, n, t, e, r, n, a, l, end subscript = Q + W) ; let’s be ‘very very’ clear of the following norms -
if heat flows out of the system, then Q will be negative
if heat flows into the system, then Q will be positive
if work is done by the system, then W will be negative
if work is done on the system, then W will be positive

Then on the hyperlink under Sawan Patel's response:

"http://physics.bu.edu/~duffy/py105/Firstlaw.html"

"The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system."

In focus, the statement of W being either negative, or positive are contradictory. I thought I understood it to be -W if the system does the work "by the system".

So, which is it? I hate for anyone else to go through my confusion. Maybe I am reading it wrong, but which one?

Great article though, it only took me five hours to go through it, work out the problems, and research the meaning of the new vocabulary words. I wish they had a computer in every classroom dedicated to the khan academy.
(1 vote)
• Can you give a better simple real-life example of the 2nd Law than that of a messy room?

The messy room is commonly used to illustrate the concept of entropy and illustrate the 2nd Law of Thermodynamics, but the simplification is egregiously misleading.

A commonly observed daily life example of something constantly moving towards a state of randomness is a ‘kid’s room’.

While the somewhat whimsical example serves its superficial purpose, it is incorrect. As long as no kid enters the room to wreak havoc, the state of the room will not tend toward randomness.

The statement:

Every time mom cleans up the room, within minutes the room looks [messy].... This is the natural state in which a kid’s room wants to exist,

is thermodynamically speaking wrong and would also violate Newton's First Law of Motion (an object will not change its motion unless a force acts on it). Once mom has put energy (W) into the system to clean up the room, it will stay in this stable condition and not, as implied, spontaneously become messy. For that to happen, you need to apply additional energy (!) in the form of work done by said kid (or an earthquake or a tornado or some other energetic disturbance, even the kid's angered mother!). In fact, you could just as easily use the same example to arbitrarily demonstrate that the Second Law is exactly the opposite, i.e., that systems tend to order if left in chaos. In both of these instances, work energy must be done to restore the room to a state of order or disorder.
(1 vote)
• I may be missing something, but where does the value for delta S in the Gibb's free energy section come from?
**delta S = -284.8 J/K***
(1 vote)
• At room temperature, if I touch a ball made of iron with one hand & a wooden ball from another hand why the iron ball feels colder than the wooden one?
(1 vote)