118 MOLECULAR MOTION AND ITS ENERGY 53 



by finding the thermal equivalent for the energy contained 

 in a gas in the form of pressure. 



In order to heat to the temperature under constant 

 pressure a mass of gas p, which at constant volume needs 

 an amount of heat 



cpS 



for the same rise of temperature, the greater amount of heat 



is needed, where C expresses the specific heat at constant 

 pressure. The difference between these magnitudes 



(C c) P 



serves therefore only to increase the volume, and thereby 

 maintain the pressure constant. 



We find the mathematical expression for the expenditure 

 of energy needed for any action, or, more shortly, for the 

 work done, by multiplying the force by the distance through 

 which it has been overcome. In our case the force on unit 

 area is given by the pressure^?, and the action consists in 

 overcoming this pressure through the whole extent of the 

 volume v, and, since the latter is assumed equal to unity, 

 the work done is 



pv = p. 



The production of this work, the initial value of which 

 was zero at the temperature of absolute zero, is equivalent 

 to the above heat, as is expressed by the equation 



p = J(C- 



J here again representing the mechanical equivalent. 



If we apply this formula, which is just the same as that 

 employed by J. B. Mayer for his calculation of J, to the 

 value of the molecular and total energies K and H, we get 

 the equation found by Clausius, viz. : 



X 

 H 



which shows that the ratio of the two energies is deter- 

 mined by that of the two specific heats. 



