The principles of thermodynamics are summarized in the form of four laws known as zeroth, first, second, and third law of thermodynamics.

It does not deal with the microscopic constituents of the matter. It deals with various macroscopic variables like pressure, temperature etc. The results of thermodynamics are useful for other branches and fields of physics and engineering like mechanical, chemical, physical, biomedical etc.

Thermodynamics is a macroscopic science which studies various interactions amongst energy, notably heat and work transfer, with matter that brings about significant changes in the macroscopic properties of a substance that are measurable. It is basically a phenomenological science based on certain laws of nature which are always obeyed and never seen to be violated.

Thermodynamics is a macroscopic science which studies various interactions amongst energy, notably heat and work transfer, with matter that brings about significant changes in the macroscopic properties of a substance that are measurable. It is basically a phenomenological science based on certain laws of nature which are always obeyed and never seen to be violated.

These are the basic laws of thermodynamics :

**$P$ =** $\frac {W}{t}$

Here, P = power, W = work,

Hence,

**$P$ = **$\frac {mgH}{t}$

The fundamental equations of thermodynamics are first and second law of thermodynamics. Both of them are combined to form a common thermodynamic relation.

**Entropy**. The energy that is not available for work is determined by using
entropy. According to the second law of thermodynamics, the entropy of the system
is constant. In a given state of equilibrium, the system has a definite value of entropy. If the system has a temperature T (in absolute scale) and a small amount of heat $\Delta$Q is given to it, we define the change in the entropy of the system as

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### Solved Examples

**Question 1: **A 0.2 m^{3} container holds oxygen at 70°C and 25 bars. Calculate the amount of oxygen in the container if the atmospheric pressure is 1 bar.

** Solution: **

$\bar{R}$-Universal gas constant, M-Molecular weight

We have $PV = nRT$

P = Pressure = 25 + 1 = 26 bars

V = Volume = 0.2 m^{3}

$R$ = $\frac{\bar{R}}{M}$

= $\frac{(8314.3 J/kg.mol-K)}{(32 kg/kg.mol)}$

= 259.8 J/kg-K

T = 273 + 70 = 343K

$m$ = $\frac{PV}{RT}$

= $\frac{26 \times 10^{-5}(N/m^{2})\times0.2 m^{3})}{(259.8 \times343 K}$

= 4.6337928 kg

**Question 2: **Given the mass of the substance is 12 grams. 3 joules of heat energy is required to raise it temperature by 6 °C. Find the specific heat capacity of the substance? [Hint:S = $\frac{1}{m}\frac{\Delta Q}{\Delta T}$]

** Solution: **

Given m = 12g = 0.012 kg

$\Delta Q$ = 3 joules

$\Delta T$ = 6 °C

S = $\frac{1}{m}\frac{\Delta Q}{\Delta T}$

= $\frac{1}{.012}\frac{3}{279}$

= 41.6666 J/kg K

### Practice Problems

**Question 1: **Suppose a man standing on top of a tower, 200m tall. He is holding a small, heavy ball that weighs 5kg. What would be the velocity of the ball be when it hits the ground assuming it was dropped on the ground with zero initial velocity? What would be the velocity of the ball be when it hits the ground assuming it was dropped on the ground with 20m/s?

**Question 2: **Water is flowing at the rate of 2 litres per minute from a tap and a geyser is heating it such that the temperature rises from 23 to 77 degree centigrade. If the geyser is supplied and aided with a burner than find the rate of combustion of the fuel given the heat of combustion is 2 $\times$ 100000 J/g.

**Question 3: **Why does the air pressure of the tire rises during driving?

**Question 4: **Why the coolant used in nuclear plants should have more specific heat?

These are the basic laws of thermodynamics :

**Zeroth law of thermodynamics****First law of thermodynamics****Second law of thermodynamics****Third law of thermodynamics**

Here, P = power, W = work,

Hence,

The fundamental equations of thermodynamics are first and second law of thermodynamics. Both of them are combined to form a common thermodynamic relation.

$dU = TdS – pdv + \sum_{i=1}^{k}u(i)dN(i)$

There are K+2 dimensions for thermodynamics space. U, S, V and N are the extensive quantities. The connections between the state variables are called equation of thermodynamic state. When the system reaches the state of thermal equilibrium then it is described with certain measurable variables that are called state variables. For example for ideal gas we write,** $PV = nRT$**

- The Evaporation of sweat from your body is an example of thermal equilibrium in action.
System : The sweat

Solution A:

Surroundings :Your body + the rest of the universe

q > 0 so, Heat flows into the system (sweat) from you in order to raise the kinetic energy of the sweat molecules enough to allow them to go from the liquid phase to the gas phase.**Solution B**

System: You

Surroundings: The sweat + the rest of the universe

q < 0 : Heat flows out of the system (you) into the sweat.

Since heat leaves your body this cools you down. This is the reason for our sweating. - Let us consider two beakers full of water. Then for one beaker, the temperature of water is above the normal room temperature, and for the other beaker it is below the normal room temperature. They are left on the table for sometime such that they both are not in contact with each other. If check the beakers after some time, equilibrium for both the beakers is reached. As observed, both the beakers of water are at the same temperature. The two beakers actually come in thermal equilibrium with the surroundings. Hence, they are in thermal equilibrium with each other also and they are at the same temperature.

If the container of the system in which the process is taking place has thermally-insulated walls or the process completes in a very short time period, so that, there is no opportunity for significant heat exchange then we say an adiabatic process has occurred. In an adiabatic process neither the heat is input nor is it released.

An adiabatic process is defined as one with no heat transfer into or out of a system, dQ = 0. We can prevent heat flow either by surrounding the system with thermally insulating material or by carrying out the process so quickly that there is not enough time for appreciable heat flow. From the first law, we find that for every adiabatic process,

An adiabatic process is defined as one with no heat transfer into or out of a system, dQ = 0. We can prevent heat flow either by surrounding the system with thermally insulating material or by carrying out the process so quickly that there is not enough time for appreciable heat flow. From the first law, we find that for every adiabatic process,

$dQ = dv + dw = 0, du = -dw$

$U_{2}-U_{1}$=$\Delta U$ = $\Delta - dW$ (Adiabatic process)

An isothermal process is a constant-temperature process. For a process to be isothermal, any heat flow into or out of the system must occur slowly enough so that thermal equilibrium is maintained. In general, none of the quantities $\Delta U$, Q or dW is zero in an isothermal process.

In some special cases, the internal energy of a system depends only on its temperature and not on its pressure or volume. The most familiar system having this special property is an ideal gas, which we will discuss in the next section. For such systems, if the temperature is constant, the internal energy is also constant; $\Delta U$=0 and Q=W. That is, any energy entering the system as heat Q must leave it again as work W done by the system

**A hypothetical setup for studying the behavior of gases**

$\Delta S$ = $\frac {\Delta Q}{T}$

In general, the temperature of the system may change during a process. If the process is reversible, the change in entropy is defined as

$S_{f} -S_{i}$ = $\int_{i}^{f} \frac {\Delta Q}{T}$

In an adiabatic reversible process, no heat is given out to the system. The entropy of the system remains constant in such a process. Entropy is related to the disorder in the system. Thus, if all the molecules in a given sample of a gas are made to move in the same direction with the same velocity, the entropy will be smaller than that in the actual situation in which the molecules move randomly in all directions.

An interesting fact about entropy is that, it is not a conserved quantity. More interesting is the fact that entropy can be created but cannot be destroyed. The second law of thermodynamics may be stated in terms of entropy as "It is not possible to have a process in which the entropy of an isolated system decreases."

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The total energy of a thermodynamic system can be measured by using enthalpy. This includes the internal energy (the energy required to create a system), and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure.

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According to the kinetic theory of gases, a gas has a large number of small particles viz atoms or molecules all of which are in constant, random motion. These moving particles collide with each other and with the walls of the container on a constant basis. The kinetic theory explains the macroscopic properties of gases, such as pressure, temperature, or volume is explained by considering their molecular composition and motion.

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According to the particle theory there is a random movement between particles in a fluid or in a gas. To describe this pre assumed movement we use the concept of Brownian movement. The Jittery motion of particles present in the liquid or gas is called Brownian motion. It is also called Pedesis. It is basically the random drifting of particles in all the directions. The particles are basically suspended in the liquids or gases. This is presumed because there is no relative motion if solids are considered.
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The table describing various thermodynamic properties of the compounds is called thermodynamics table. An example is shown below:The following problems help us to learn more about this topic :

$\bar{R}$-Universal gas constant, M-Molecular weight

We have $PV = nRT$

P = Pressure = 25 + 1 = 26 bars

V = Volume = 0.2 m

$R$ = $\frac{\bar{R}}{M}$

= $\frac{(8314.3 J/kg.mol-K)}{(32 kg/kg.mol)}$

= 259.8 J/kg-K

T = 273 + 70 = 343K

$m$ = $\frac{PV}{RT}$

= $\frac{26 \times 10^{-5}(N/m^{2})\times0.2 m^{3})}{(259.8 \times343 K}$

= 4.6337928 kg

Given m = 12g = 0.012 kg

$\Delta Q$ = 3 joules

$\Delta T$ = 6 °C

S = $\frac{1}{m}\frac{\Delta Q}{\Delta T}$

= $\frac{1}{.012}\frac{3}{279}$

= 41.6666 J/kg K

More topics in Thermodynamics | |

Phase Change | Laws of Thermodynamics |

Adiabatic Process | Isothermal Process |

Entropy | Enthalpy |

Brownian Motion | Kinetic Theory |

Gas Laws | |

Related Topics | |

Physics Help | Physics Tutor |