A9 m3 container is filled with 300 kg of r-134a at 10°c. what is the specific enthalpy (kj/kg) of the r-134a in the container? (4 significant figures)

Given the mass of R-134a m = 300kg; Volume of the container V = 9  cu. meter; Temperature of R-134a T = 10 degrees Celsius; 
Formula of specific volume : v = V / m = 9 / 300 = 0.03 cu. m / kg. 
At T = 10 degrees Celsius from saturated R-134a tables, vf = 0.0007930 cu. m /kg; vg = 0.049403 cu. m/kg. We know v = vf + x (vg - vf), so 0.03 = 0.0007930 + x (0.049403 - 0.0007930), which makes x = 0.601.  
Specific enthalpy of R-134a in the container is h = hf + x*hfg = 65.43 + (0.601 * 190.73). Answer is 180.0587 kJ/kg

Amagnetic field "does not" interact with stationary electric charges. so, a.hope my answer you : )