Fig. Joule’s apparatus |

I hope you understand the concept of Joule’s law for perfect gas, still any query write to us in comments section.

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If a fixed mass of gas is allowed to expand without doing external work under the condition that no heat enters or leave the gas, i.e. the gas is allowed to undergo free expansion, then If the molecules of the gas attract one another, internal work will have to be done by the gas in pulling them apart and this will happen at the cost of internal kinetic energy of the gas which will therefore be cooled by the expansion.

On the other hand If the molecules repel one another the expansion should be accompanied by a rise in temperature and if there is no intermolecular attractions, then there should be no change in temperature.

Joule’s Experiment:-

Joule took two Copper cylinders A and B joined by a tube fitted with a stop cock. A contains air at high pressure while B is perfectly evacuated. Both cylinders were immersed in the minimum amount of water, whose tempreture was noted with a sensitive thermometer.

Fig. Joule’s apparatus |

The stop cork was opened and air from cylinder A expanded into the cylinder B(vacum) and the expansion was so rapid that no heat communication was possible with the surrounding and hence a cooling was expected if intermolecular attractions exists

but Joule could not detect any change in tempreture.

Then he changes its experiment by emerging A and B in separate water baths but still no trace on intermolecular attractions was observed i.e. the fall of tempreture in A was exactly equal to rise in temperature of B.

If however intermolecular attractions existed then the fall in temperature in A would have been greater than the rise in temperature of B.

Hence Joule’s law stated that the Internal energy of a gas depends only on its temperature and is independent of its volume & pressure.

I hope you understand the concept of Joule’s law for perfect gas, still any query write to us in comments section.

Suppose Q is the heat given to a body and its temperature increases by ΔT, then Heat capacity of body is given by :-

See Also:- (i) Heat Internal Energy & Work

(ii)Work done during Adiabatic Process

Further, Heat Capacity per unit mass is known as Specific Heat Capacity & it is denoted by ‘C’

Thus, __” Specific heat of the material is quantity of heat required to raise the temperature of unit mass by 1°C or 1°K. “__

See Also :

(i) Work done in Isothermal Process

(i) Work done in Isothermal Process

(ii) Slope of Adiabatic Curve is more steeper than Isothermal.

As a gas is compressible, then there is a rise in temperature without supplying any heat to the gas, then,

On the other hand, if heat is supplied to gas and gas is allowed to expand such that there is no rise in temperature then,

See Also : (i) Quasi-Static Proces

(ii) Adiabatic Process & its equation

(iii) Isothermal process & its equation

(ii) Adiabatic Process & its equation

(iii) Isothermal process & its equation

Hence In order to find the value of Specific heat of gas i.e. the pressure or volume has to be kept constant.

Hence we have two specific heats :-

It is defined as *“the amount of heat required to raise the temperature of unit mass of gas through 1°C, when its volume is kept constant”.*

It is denoted by Cv.

It is defined as *“the amount of heat required to raise the temperature of unit mass of gas through 1°C, when its pressure is kept constant”.*

It is denoted by Cp.

The unit of Specific Heat in C. G. S system is cal/gm/°C.

When a gas is heated at constant pressure, it expands. Therefore heat supplied at constant pressure is partially used in doing work & remaining in raising the internal energy of the gas.

On the other hand, when a gas is heated at constant volume no work is done and hence total heat supplied is used to rise the temperature of the gas.

Thus, more heat is required for increasing the temperature of the gas through 1°C at constant pressure then at constant volume.

Consider two isothermals AB & CD drawn for 1 mol of ideal gas at closed temperature T & T+ΔT.

Fig. Isothermals AB & CD |

When gas expands at constant volume, then change in internal energy is given by :

When gas expands at constant pressure, then change in Internal Energy :

Now, From Ideal gas equation:

Now,

I hope you’ll understand Concept of Specific Heat & Mayer’s Formula. For any query or suggestions ,write to us in comment section.

Here are some questions which can easily explain all the basics of the first law of thermodynamics:

According to this law,

The total energy supplied to the system is spent partially in increasing internal energy of the system & the rest of the energy is spent in doing work on the surrounding.

Mathematically it is written as :

where,

ΔQ = total energy supplied(heat)

ΔU = change in internal energy

ΔW = work done on the surrounding.

ΔW = work done on the surrounding.

Here Internal Energy is a state function. Hence it is perfect differential, while Heat & Work are path function, hence these are not perfect differential.

**See Also :**

Let us suppose that a quantity dQ of heat is supplied to a body. It is in general spent in three ways :

If dUk & dUp are respectively the change in internal kinetic energy & internal potential energy & dW be the external work done then since *“energy can neither be created nor destroyed but only convert from one form to another”,*we have

dQ = dUk + dUp + dW

But dUk + dUp = dU, the increase in total internal energy of the body.

Hence, dQ = dU + dW

or

This equation represent the differential form of the first law of thermodynamics, which may therefore be stated.*“In all transformations, the energy due to heat units supplied must be balanced by external work done plus an increase in internal energy.”*

Thus, the first law holds for the law of conservation of energy.

See Also :

The first law of thermodynamics establishes an exact relationship between heat & work. According to it a definite quantity of heat will produce a definite amount of work and vice versa. It denies that work or energy can be created out of nothing. It means that it is impossible to construct a thermal machine which may operate without any expenditure of fuel and may thus create energy out of nowhere. A machine that would do this could run itself and is generally called a perpetual motion machine of the first kind.

The first law rules out such mechanics and is therefore sometime defined in the form,

“Perpetual motion machine of the first kind is impossible “.

Three ideas are included in the first law of thermodynamics :

See Also :

More Related Posts :

- Isothermal Process & Isothermal equation of perfect gas.
- The equation of state(Relation between P, V & T)

- Thermodynamic Process
- Thermal equilibrium & Zeroth’s law

Hope you understand the First law of thermodynamics, still, any query. Write to us in the comment section.

See Also:

- Thermodynamics & thermodynamic systems
- Joule-Thomson Expansion: Porous Plug Experiment
- Specific Heat of Gas (Relation b/w Cp & Cv)
- Heat Engine & It’s Components
- Heat, Internal Energy & Work – Thermodynamics
- Joule’s Law for a Perfect Gas
- Vanderwall’s Gas in Joule Thomson Expansion-Thermodynamics
- Thermal Equilibrium & Zeroth’s law of thermodynamics

Heat is defined as flow of energy from one point to another.

Thermodynamically heat is defined as flow of energy from one body to another on account of difference in temperature.

♦ If the system absorbs the heat then heat is taken as positive, while of system releases the heat then heat is taken as negative.

♦ Heat is path dependent.

The system has some amount of energy which is not apparently shown by the system but is capable of doing work. Such type of energy is known as Internal Energy or Intrinsic Energy.

According to Kinetic theory, all the matters are made up of large number of tiny particles, which are in continuous movement inside the body.

The energy due to movement of particles inside the body is called molecular Kinetic Energy, & the energy due to molecular attraction between the particles is called molecular potential energy.

Here, Internal Energy is denoted by U, molecular Kinetic energy is denoted by Uk & molecular potential energy is denoted by Up such that :-

U = Uk + Up

♦ Internal Energy is a state function.

♦ Internal Energy of a body is depend on initial & final position & does not depend on path followed.

U = Uf – Ui

♦ Internal energy depends upon temperature.

♦ In a cyclic process, total internal energy of a complete cycle is zero.

Work is defined as the way by which system may exchange heat from surrounding.

If force F produces displacement dx, then the work done by the system :-

I hope this tutorial is helpful for you to understand Heat, Internal Energy & Work. Still any question you can ask us in comment section.

Today we’re going to find work done during an adiabatic process. For this, Let us consider one gm. a mole of perfect gas enclosed in a cylinder having perfectly insulated walls.

Let it expands adiabatically from an initial volume V_{1} to a final volume V_{2}. Then work done by the gas during the expansion is given by :

where P is the instantaneous pressure of the gas while suffering infinitesimal expansion dV.

Now during an adiabatic process:

Hence, work done in the adiabatic process depends only upon initial & final temperature T_{1} & T_{2.} Thus the work done along any adiabatic between two isothermal is independent of particular adiabatic.

I hope this article helps you.

If you need more help on this topic (**Work done during an Adiabatic Process**), feel free to drop your questions in the comment section.

- Show that the Slope of Adiabatic Curve is steeper than that of Isothermal Curve?
- Reversible and Irreversible Processes (Thermodynamic Processes)
- The Equation of State (Relation between P, V, and T )
- Adiabatic Process & Adiabatic Equation for a perfect gas
- First Law of Thermodynamics & Its limitations
- Vanderwall’s equation of State- Thermodynamics
- Heat Engine and Its Components
- The expression for Joule’s coefficient – Thermodynamics
- Zeroth law of thermodynamics & Thermal Equilibrium
- Thermodynamics & thermodynamic systems

If A is the area of cross section of the piston, then force exerted by the gas on the piston is :

F = P.A

If the piston displaced outward through small displacement dx, then the small amount of work done:

dW = F.dx

or dW = P.(A.dx)

or dW = P.dV

Therefore, the total work done by the gas in expansion from V1 to V2 is given by :

& total amount of heat absorbed can be calculated as:

Q = amount of heat absorbed (calories).

- Reversible and Irreversible Processes (Thermodynamic Processes)
- Isothermal Processes & Isothermal Equation for a perfect gas
- Expression for Joule’s coefficient – Thermodynamics
- Work done during an Adiabatic Process
- First Law of Thermodynamics & Its limitations
- Show that the Slope of Adiabatic Curve is steeper than that of Isothermal Curve?
- Specific Heat of Gas (Relation b/w Cp & Cv)
- Joule’s Law for a Perfect Gas
- Quasi-Static Process (Nearly Static Process)
- Vanderwall’s equation of State- Thermodynamics
- The Equation of State (Relation between P, V, and T )

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