 Welcome back. Now let us combine the results which we have obtained of Hylako and obtain an expression for the efficiency of a Carnot cycle which is running using an ideal gas with constant specific heats as its working fluid. And you should remember that we used I-DL gas with a constant specific heats as the working fluid just to simplify the associated algebra and calculus. Let us obtain an expression for the efficiency of this cycle. Let us call it eta C, C for Carnot. The efficiency of our Carnot cycle will be 1 minus heat rejected divided by heat absorbed. And this will be 1 minus expression for heat rejected is MRTB ln V2 by V3 divided by expression for Q absorbed MRTA logarithm of V1 by V4. It is obvious that mass and the gas constants cancel out from the numerator and denominator. Let us see what we can do with this particular term. While analyzing the two adiabatic processes, we have noted that we obtain this expression T1 V1 raise to gamma minus 1 is T2 V2 raise to gamma minus 1 and T4 V4 raise to gamma minus 1 is T3 V3 raise to gamma minus 1. Now, from this you will notice that we have from this V2 by V1 raise to gamma minus 1. V2 by V1 raise to gamma minus 1 will be T1 by T2. But T1 equals TA, T2 equals TB, TA in turn equals T4 and TB in turn equals T3 and from this expression we have this equal to V3 by V4 raise to gamma minus 1. This indicates that we must have V2 by V1 equal to V3 by V4 or by transposing terms V2 by V3 equal to V1 by V4 and that means because these ratios are equal, the value of this sub expression equals 1 and that means the Carnot cycle has an efficiency equal to 1 minus TB by TA. Now, this is an expression obtained using our analysis and here TA, TB are temperatures on the ideal gas Kelvin scale. But we also have this expression, efficiency of the same cycle will be theta B by theta A. Why? This is because the Carnot cycle is running an engine which is working as a reversible 2 T heat engine and by the second law of thermodynamics and Carnot theorem and our definition of the thermodynamic temperature scale, the efficiency of any such machine must be 1 minus theta B by theta A and this is from the second law, Carnot theorem and definition of thermodynamic temperature scale. So, what we have been able to show using this expression, we have been able to show that theta B by theta A equals TB by TA. That means the ratio of temperatures pertaining to two isotopes is the same whether the two temperatures are measured on the ideal gas Kelvin scale which is the right hand side of this expression or the thermodynamic Kelvin scale which is the left hand side of this expression. Transposing terms, this implies that theta B by TB is theta A by TA which indicates that the thermodynamic temperature scale and the ideal gas temperature scale, the Kelvin scales, the temperatures are proportional to each other. And now let us take one particular isotherm that at the triple point of water and that would be theta ref by T ref and we have defined both of these to be the same value 273.16, 273.16 both on the Kelvin scale, one thermodynamic and one ideal gas. And since this equals one, we have been able to show that the temperature measured on the thermodynamic Kelvin scale equals the temperature measured on the ideal gas Kelvin scale. What is the advantage of this? Remember that measurement of the thermodynamic temperature requires an ability to implement a 2T heat engine which we know is impossible in practice. By showing and demonstrating that the temperatures on the thermodynamic Kelvin scale will equal the temperature on the ideal gas Kelvin scale, we can measure the temperature using the ideal gas Kelvin scale. And ideal gas can always be realized to a very good approximation by working with a real gas at low pressures and low density. Thank you.