Answer to Question #105412 in Molecular Physics | Thermodynamics for Kule

Question

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

Answer 1

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 + $\frac{400\times0.518}{5.403}$ =134.10 BTU/lb

b)

change in enthalpy =

$\Delta$H=p₁V₁ - p₂V₂ =0

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