Question
please explain how
The piston diameters in Fig. P3-119 are D1 = 10cm and D2 = 4 cm. Chamber 1 contains 1 kg of helium, chamber 2 is filled with condensing water vapor, and chamber 3 is evacuated. The entire assembly is placed in an environment whose temperature is 200degreeC. Determine the volume of chamber 1 when thermodynamic equilibrium has been established. Answer: 3.95 m3 At equilibrium, the forces should get balanced. Hence the force (F1) on the piston due to pressure of first chamber should be balanced with the force (F2) on the piston due to pressure of second chamber. The third chamber is evacuated and thus it doesn't exert any pressure (or force on piston). The second chamber contains water at 200degreeC. Hence pressure that exists in second chamber is the saturation pressure corresponding to 200 degree C P2 = Pat_200degreeC +1555kPa Balance the forces acting on the piston F1 = F2 P1A1 = P2A1 P1times pi / 4 = P2times pi 4 P1 = p2times Here the diameters of chamber 1 and 2 are represented as D1 and D2. Now substitute 10 cm for D1. 4 cm for D2. and 1555 kPa for P2 in above equation. P1 = P2 times = 1555 times 42 / 102 = 248.8kPa Use the ideal gas equation and find the volume of first chamber, as the first chamber is filled with helium. V1 = mRT / P1 Substitute 248.8 kPa for p1, 1 kg for m, 2.0769 kJ/kg K for R, and (200+273) K for T V1 = mRT / P1 =1 times 2.0769 tiems (200 + 273) / 248.8 = 3.95 m3 Thus the volume of first chamber is gases such as carbon dioxide can be treated as ideal gases with negligible error (often less than 1 percent). Dense gases such as water vapor in steam power plants and refrigerant vapor in refrigerators, however, should not be treated as ideal gases. Instead, the property tables should be used for these substances
Explanation / Answer
If the pressure of the water is below 10 kPa then water vapor can be treated as an ideal gas.
but if the pressure is higher, the error will be unacceptable.
Moreover Between the two states is a gray area. In that case you should look at the compressibility factor, Z=Pactual/Pideal. Z is a function of reduced pressure Pr and reduced temperature Tr (more on these later), and this correlation is given in standard charts which apply for most substances . If you accept errors up to 10%, you may apply the ideal gas law as long as 0.9<Z<1.1. So:
Pure water vapor at 1 atm and 373 K has Pr=1/217=0.0046, so the ideal gas law applies to within 10% error.
So the ideal gas law applies . But these rules only apply if you accept errors up to 10%. If accuracy is important, only use the ideal gas law for Pr<0.025 and don't use it for saturated vapors at all. When the ideal gas law doesn't apply, correct it using the compressibility factor (Pactual=ZPideal) or use a better equation of stat
