That which follows the simple equation is called ideal gas.

PV= Mrt                                                          … (2.9)

here P= pressure, V= volume and T= temperature, m= mass, R= gas constant of the gas. Now 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.

However, inreality, we can never find any gas similar to ideal gas. However, all the gases tends to behave like an ideal gas at low pressure and density.

An ideal or perfect gas has mass m which transforms from P1, V1, T1 to P2, V2, T2 then from equation (2.9) we can get is,

P1 V1 / T1 = P2 V2 / T2 = Mr = constant

T2 / T1 = P2 V2 / P1 V1

Theequation(2.10) mightbe utilised for computing or relating the temperatures. The correlationbetween the two equations are given in Equation (2.9) is what called ideal gas equation of thestate.

Solved example

Example 2.1. The temperature t (0c) is displayed on a thermometer andis definedregarding a property K by the relationt =aInK+ b.

Where a and b are constants. The constant K is said to be 1.83 and 6.78 at the ice point and the steam point at the temperature scale where 0 and 100 are allottedrespectively. Now determine the temperature equivalent to K which equals to 2.42 on the standardthermometer.

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

Now 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 = 21.34oc(Ans)

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