An ideal gas is one which follows the following simple equation.

PV= Mrt … (2.9)

where P, V and Tare the pressure, volume and temperature of the gas having mass m, and R is the gas constant. Dividing both sides of equation (2.9) by m we get,

P ( V / m) = RT

Pv = RT where v = V/ m = specific volume of the gas.

In reality, there is no gas which is an ideal gas or perfect gas. However all gases, at a very low pressure and density approach to ideal gas.

If the state of an ideal or perfect gas having mass m is changed from P_{1}, V_{1}, T_{1} to P_{2}, V_{2}, T_{2 }then from equation (2.9) we get

P_{1 }V_{1} / T_{1} = P_{2} V_{2} / T_{2} = Mr = contant

T_{2} / T_{1} = P_{2} V_{2} / P_{1} V_{1}

The relationship of eqn. (2.10) may be used for calculating or comparing the temperatures. The relationship given in equation (2.9) is called ideal gas equation of state.

**SOLVED EXAMPLE**

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**Example 2.1.** The temperature t (0c) on a thermometric scale is defined in terms of a property K by the relation

t =a InK+ b.

where a and bare constants. The values of K are found to b~ 1.83 and 6.78 at the ice point and the steam point, the temperature of which are assigned 0 and 100 respectively. Calculate the temperature corresponding to a reading of K equal to 2.42 on the termometer.

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**SOLUTION. ** t = a In K + b

K = 1.83 K = 6.78

t = 0 t = 100

0 = a In 1.83 + b

100 = a In 6,78 + b

Solving (2) and (3) for a and b we get

a = 76.355 , b = – 46.14

t = 76.355 In k – 46.14

k = 2.42 => t = 21.34^{o}c.Ans