Question #105412

a reversible non flow process of a perfect gas proceeds from a pressure of 400 psia to a pressure of 100 psia with a corresponding increase in specific volume of the gas from 0.518ft^3/lb to 2.072 ft^3/lb. during the process the internal energy remains constant at 95.76 BTU/lb. calculate the: a. the enthalpy of the gas at 400 psia and 0.518 ft^3/lb in BTU/lb b. the change of enthalpy between the initial and final states

Expert's answer

As the internal energy is constant here we can assume it as isothermal process where the temperature is constant

a) Formula for enthalpy is :

H=U+pV

H=95.76 BTU/lb + 400×0.5185.403\frac{400\times0.518}{5.403} =134.10 BTU/lb


b)

change in enthalpy =

Δ\DeltaH=p1V1 - p2V2 =0

Here change in enthalpy is equal to zero as the internal energy is constant and it is a isothermal process




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