Answer to Question #225643 in Molecular Physics | Thermodynamics for eren

Question

R-134a flows in a pipe at 30^oC with a specific volume of 0.04434 m³/kg. The internal energy and enthalpy of R-134a are 250.8 kJ/kg and 273.0 kJ/kg, respectively. Determine the pressure of the refrigerant in MPa. in 2 decimal places

Answer 1

$V=0.04434 \;m^3/kg \\ U = 250.8 \;kJ/kg \\ h= 273.0 \;kJ/kg \\ h=U+pV \\ 273.0 = 250.8 + P \times 0.04434 \\ P = 500.67 \;kPa = 0.50 \;MPa$