# Ideal Gas And Heat Engine

Gases are made up of many particles that move randomly and the particles interact with each other. Ideal gas does not exist in reality and is a theoretical gas. An ideal gas is a condition in which every collision between molecules or atoms are absolutely elastic and there does not exist intermolecular attractive forces. Hence, the internal energy of the system is in the form of kinetic energy and any change in internal energy is followed by a change in the temperature.

From the gas laws, the ideal gas can be described by

Pressure (P)

Volume (V)

Temperature (T)

As we know, for an ideal gas

PV = K

$\frac{V}{T} = k$

Then,

$\frac{PV}{T} = constant$

Hence, the ideal gas equation is given by

PV = nRT

where,

P= pressure of the gas;

n= Number of Moles;

V=volume of the gas;

T=Absolute temperature;

R=Ideal Gas constant (Boltzmann Constant)

Using the ideal gas equation, the behavior of any gas can be studied under STP condition.

Note: One mole of an ideal gas at STP occupies 22.4 liters.

Now let us learn about the heat engine and the efficiency of the heat engine.

## The Efficiency Of Heat Engine

The Heat engine is a device that converts heat energy or chemical energy into mechanical energy that is used to do work. The advantage of using heat engine is that most of the energy can be efficiently transformed to heat by processes like exothermic reactions, absorption of light or energetic particles, resistance, dissipation, and friction. Heat engines are versatile and can be used in various fields.

The efficiency of the heat engine is represented by ‘η’

The efficiency of the heat engine is given by the ratio of the output of work to the heat supply of the heat engine. Given by

$\eta = \frac{W}{Q_{1}}$

Where,

W is the work done

Q1 is the heat absorbed from the source.

Note: The internal energy of the system is not affected since the engine returns to its original state after the completion of each cycle.

Hence,

$\Delta U = 0$

W = Q1-Q2

The efficiency is given by

η =(Q1−Q2)/Q1

$\eta = 1 – \frac{Q_{1}}{Q_{2}}$

When Q2 = 0, the efficiency of the heat engine will be 100%. Practically this is not possible since there will be some loss in the system.

For Carnot heat engine (reversible engine) the efficiency will be maximum.

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