In this article, MechStudies tries to explain all about the first law of thermodynamics like definition, how it’s related to the conservation of energy, identifying instances where the law works in everyday situations, and finally in calculating changes in the internal energy after considering the work done and heat transferred.

So without any further adieu let’s begin!

Before going to the discussion about the first law of thermodynamics, let’s try to understand the basics of thermodynamics.

Whether you’re fuelling up to drive a car or having lunch, you’re always trying to convert energy from fuel to run the car and from food to function your body.

To understand how this works we need to learn about thermodynamics and the laws behind it.

Energy is never constant it is always being converted from one form to another.

- Let’s take an example of food, we eat and utilize the energy stored in it so that we can use it to power our body instantly or just store it in the form of fats for future use.
- Not all energy conversions happen at such a small scale, many engineering applications use energy conversions at a large scale like the hydroelectric power plant.
- In hydroelectric dam potential energy of the water is converted into kinetic energy to move the turbines which then turns the shaft connected to the electric generator, thus converting the movement of water to electrical energy.

Energy cannot be created out of anything thus these conversions are important so that the required form of energy is derived from some other type of energy. Thus thermodynamics is essential to know how energy can be converted.

Thermodynamics is the branch of physics and engineering that deals with the conversion of energy from one form to another often in the form of heat and work.

- It also gives insights into how thermal energy can be converted to and from other forms of energy and also work.
- Thermodynamics is the major focus area for mechanical engineers as they need to know how much heat or work they will get out of a system if they put energy into it.
- It also finds its purpose in chemical energy as during the formation of new compounds energy is released due to the chemical reactions often that energy is thermal energy.

Let’s try to understand the first law of thermodynamics,

- When a system completes one cycle the net work is either done
**by**the system or**on**the system. If we consider a cycle in which the work is done by the system, in this case, some other source of energy (Thermal or electrical) must have provided the mechanical energy. - After the system comes back to its initial state the intrinsic energy of the system remained intact. Thus the only source of energy in the cycle was heat energy which was given and rejected in various stages.
- Hence by the law of conservation of energy, the net work done by the system is equal to the net heat provided to the system.

Thus the first law of thermodynamics can be stated as follows,

- There is another type of energy that is different than the other types of energy available in the system called
**Internal energy (U).**It is associated with the random movement of the molecules. - It is similar to potential and kinetic energy but on a microscopic scale. For example, water resting might seem stagnant to the naked eye but on a microscopic level, its molecules are moving at super-high speeds!
- Despite being at the microscopic scale, internal energy may have a major effect on a system because this can result in changes in temperature, changes in phase, or changes in the chemical structure.
- Just like mechanical energy is the combination of kinetic and potential energy, the internal energy is the combination of mechanical energy of its atoms and molecules.

The first law applies conservation of energy principle to the system in which heat transfer and work done are the means of transporting energy in and out of the system. The first law of thermodynamics also states that the change in the internal energy of the system is equal to the difference between the net heat transmitted into the system and the net work done by the system.

It can be written as,

**ΔU = Q − W**

- ΔU – Change in the internal energy
- Q – Net heat transferred into and out of the system
- W – Network is done by the system

- U is the function of parameters such as pressure, volume, and temperature thus we can calculate changes in its internal energy from a few macroscopic variables based on heat transferred or work done.
- Potential, kinetic, and internal energy shows us what can exist within the system but these types of energy can’t cross their boundary from their system to the surroundings.
- The main types of energy that can cross boundaries are heat (Q) and work (W).

**Let’s learn about Heat (Q) and Work (W) – **The only means of bringing in and taking energy out of the system.

- Heat transfer is done by the temperature difference created between two systems.
- Work is the transfer of energy through external macroscopic forces resulting in the change in external parameters.
- Heat and work can create similar results – for example, in a cylinder and piston arrangement when the gas is heated the temperature is increased due to heat energy transfer, if in the similar arrangement the plunger is pushed down and the gas is compressed still the temperature will increase.
- Thus once the temperature of the system increases it becomes extremely difficult to identify the cause of change – Heat or work.
- Heat and work are based on the ongoing process, for a system at a stable state heat and work have no meaning.

- Heat is the energy in transit and hence can be identified at the system boundaries and can be stored in a system.
- Being the vector quantity heat has magnitude, point of action, and direction. Heat energy is measured in kJ or BTU (British thermal unit).
- When the heat is transferred in the system increasing its energy, it means that Q is positive. When the heat is dissipated to the surroundings, in turn, decreasing the energy of the system it denotes that Q is negative.

- Work is the transfer of energy through mechanical means through interaction between the system and its surroundings.
- Work done by the system is considered positive, work done on the system is considered negative.

**Specific heat at constant volume and pressure**

The energy needed for raising the temperature by 1 degree of a unit mass of a substance is known as specific heat. The equation is given by:

**dQ = mcdT**

Where,

- m = mass of a substance,
- c = specific heat
- dT = rise in temperature

**Specific heats are of two types:**

**Cv – Specific heat at constant volume**– This is the energy required to raise the temperature of a substance when the volume is kept constant.**Cp – Specific heat at constant pressure**– This is the energy required to raise the temperature of a substance when the pressure is kept constant.

- Specific heat at constant volume is always less than the specific heat at constant pressure as the system expands at constant pressure and the energy generated due to this expansion is also supplied to the system.

For a closed system undergoing a constant volume process, let us consider this in the first law of thermodynamics in differential form:

**dq – dw = du**

No work is done under constant volume process, thus by its definition, we can write:

**Cv*dT = du**

Similarly, for a constant pressure process using the first law of thermodynamics, the equation can be written as: (In constant pressure process – wb + dU = dh)

**Cp*dT = dh**

- Cp and Cv are properties and do not depend on the type of process happening.
- Cp is based on the changes in enthalpy (h) and Cv is connected to the changes in internal energy (U)
- Thus Cp can also be defined as the change in specific enthalpy per unit temperature change of a substance at constant pressure.
- And Cv can be given as the change in specific internal energy of a substance per unit temperature change at constant volume.

So for a reversible non-flow process at the constant volume we have –

**dQ = m Cp dT**

For a reversible non-flow process at constant pressure we have –

**dQ = m Cv dT**

For a perfect gas, Cp and Cv values are constant at all temperatures and pressures for anyone gas. Hence from the above two equations, we get,

Heat flow in a reversible constant volume process

* *= mCv (T2 – T1)

Heat flow in a reversible constant pressure process

**= mCp (T2 – T1)**

In actual practice when real gas is used, the values of Cp and Cv vary with respect to temperature but an average value may be used for most of the solutions.

**Relationship between Cp and Cv:**

Let’s take gas and heat it at constant pressure from T1 to T2

Thus as per the non-flow equation,

- Q = (U2 – U1) + W

For a perfect gas,

- U2 – U1 = mCv (T2 – T1)
- Q = mCv (T2 – T1) + W

Work done by fluid in a constant pressure process is given by:

- W = PV (pV = mRT)
- W = p(V2 – V1)
- W = mR(T2 – T1)

Thus on substituting the values we get,

- Q = mCv (T2 – T1) + mR(T2 – T1)
- Q = m(Cv + R) (T2 -T1)

Now as the process is constant pressure,

- Q = mCp (T2 – T1)

Combining the two equations we get,

- mCp (T2 – T1) = m(Cv + R) (T2 -T1)
- Cp = Cv + R
- Cp – Cv = R

Dividing both sides by Cv, we have,

Cp/Cv – 1 = R/Cv

or, Cv = R/(γ-1) [Where γ = Cp/Cv]

Similarly to get the value of Cp, dividing both sides by Cp, we have

Cp = γR/(γ-1) [Where γ = Cp/Cv]

- The sum of the internal energy (U) and the product of its pressure and volume (pv) is known as the enthalpy of a system.

The equation is given as:

**h = u + pv**

- Enthalpy is similar to energy and is measured in joules or ergs. It can be determined by the values of pressure, temperature, and composition of the system.
- The change in internal energy is equal to the difference between the heat transmitted to the system and the work done by it. The enthalpy change is precisely equal to the heat transmitted to the system if the only work done is a change in volume at constant pressure.
- The amount of energy required to transform the phase of a substance from liquid to gas is known as the enthalpy (or latent heat) of vaporization and is represented in joules per mole.

**For a perfect gas,**

- h = u + pv
- h = Cv*T + RT
- h = (Cv + R)T
- h = Cp*T
- h = Cp*T

**H = mCp T**

In a constant volume process, the substance is contained in a closed chamber due to which its volume can neither be increased nor decreased. As there is no change in the volume of gas the work is done is zero on or by the system.

Thus isochoric process implies work (W) is always zero unless stated otherwise.

Applying the first law of thermodynamics and considering the mass of a substance as unity, we have

- Q = (U2 – U1) + W

But for Isochoric process, W = 0 – since change in volume (dV) = 0

- Q = (U2 – U1) = Cv (T2 – T1)

Thus for a system with mass (m)

- Q = mCv (T2 – T1)

The process in which the pressure doesn’t change is known as the Isobaric process. The boundary of the system is not rigid as in the case of a constant volume process it moves when the external force is applied to it.

A piston and cylinder arrangement can be taken as an example for a constant pressure process where the piston is pushed by the force exerted by gas, here the work is done by the gas on its surrounding.

Applying the first law of thermodynamics to the process by considering the mass of substance as unity.

- Q = (U2 – U1) + W

Work done, W = p(V2 – V1)

- Q = (U2 – U1) + p(V2 – V1)
- Q = U2 – U1 + pV2 – pV1
- Q = (U2 + pV2) – (U1 + pV1)

Since h = U + pV, we get –

- Q = h2 – h1 = Cp(T2 – T1)

Where,

- h = Specific enthalpy
- Cp = Specific heat of a substance at constant pressure

For a substance with mass, m

**Q = H2 – H1 = mCp (T2 – T1)**

Where H = mh

- The process that involves no change in the temperature of a substance is known as the isothermal process.
- Let’s take an example of piston and cylinder, in a normal case when the piston is expanded from high pressure to low pressure the temperature decreases but in the isothermal process, heat is added continuously to keep the temperature constant.
- In case of piston compression where the temperature increases heat is removed continuously from the system to keep the temperature constant.
- Constant temperature implies that the internal energy of the system is constant

Applying the first law of thermodynamics to the process by considering the mass of substance as unity.

- Q = (u2 – u1) + W
- Q = Cv (T2 – T1) + W
- Q = 0 + W

Work done is given by,

Here pv = constant, we have

C can be written as

p1v1 = p2v2 = constant,

The final equation will be,

When there is no transfer of heat in or out of the system during the process it is known as the isentropic process. It can be reversible or irreversible, considering the reversible adiabatic process for this part, we get

- Q = (u2 – u1) + W
- O = (u2 – u1) + W
- W = (u1 – u2)

The equation of work done is valid for both reversible and irreversible adiabatic processes. Perfect thermal isolation is a prerequisite for an adiabatic process to take place.

**Deriving an equation for work done in adiabatic**

- As there is no transfer of heat between the system and the surrounding, dQ = 0
- dU + PdV = 0
- Cv dT + P dV = 0
- Cv dT + RT/v . dV = 0
- dT/T + (R/Cv). dV/V = 0
- dT/T + (γ-1) dV/V = 0

Integrating the above equation we get,

**TV ^{(γ-1) }= constant**

**What is adiabatic expansion?**

When external effort is applied to a system, utilizing the internal energy of the gas, resulting in a reduction in the temperature of the gas, this is referred to as adiabatic expansion.

**What is adiabatic compression?**

Adiabatic compression is described as the compression of air in which no heat is generated or removed from the air and the internal energy of the air is raised to match the external work done on the air. As the temperature rises during compression, the pressure of the air exceeds the volume.

The polytrophic process is defined as a reversible process for any open or closed system consisting of gas and vapor that involves both heat and work transfer while maintaining a certain combination of properties constant.

In many non-flow processes, vapors and perfect gases obey the law closely.

For a reversible process, we have

**W =**** ∫p dV**

For polytropic process, we have, **PV ^{n }= Constant**

p = C/V^{n}

Thus the work in the polytropic equation is given by,

Thus the work in the polytropic equation is given by,

- It is a process in which no work is done on or by the fluid and no heat is transferred in or out of the system. Free expansion is similar to the adiabatic process but completely irreversible.
- To visualize the process let’s consider two containers thermally insulated connected through a short conduit. Let the first container be filled with compressed gas and the other one be completely vacant. When the first container is opened through a valve, the stored pressurized gas will be expanded to the other container. Now the overall pressure will be reduced as compared to the initial pressure.

Applying the first law of thermodynamics for the initial and final state of the system,

Q = (u2 – u1) + W

As no work is done by or on the system since the system is rigid, the total work done will be zero. The system is thermally insulated thus the heat flow will be equal to zero.

- u2 – u1 = 0
- u2 = u1

Thus in free expansion internal energy remains constant.

U = CvT

Thus the free expansion of a perfect gas is given by,

- CvT1 = CvT2
- i.e. T1 = T2

For a perfect gas undergoing free expansion – the initial and final temperature of the system is equal.

Hence, we have got a basic idea about the first law of thermodynamics, its applications. Any doubt, please let us know. Thanks

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