Carbon dioxide is compressed in a tank to a pressure of 100.0 atm. and a tempera
ID: 3666399 • Letter: C
Question
Carbon dioxide is compressed in a tank to a pressure of 100.0 atm. and a temperature of 300K. Find the volume of the gas by iterating the van der Waal equation starting with a value for v found from the ideal gas law. Substitute this value for v into the v2 term and solve for the v term. Then use this new value for v to go back and compute v2 and get a new value for the linear term in v, etc. The van der Waal constants for CO2 are a = 366 x 103 J m3 kmole-2 and b = 0.0429 m3 kmole-1. Write a FORTRAN program that does this. The answer you get should be 0.080 cubic meters per kilomole. How many iterations does it take for the answer to converge to this number? Iterative solutions for complex equations are a common practice in physics. Print out a file with the following: On the first line should be the initial value of v computed from the ideal gas law, identified as such, then 2 columns with the iteration number in the first column and value of v computed from the iteration. It should look like the following:
Explanation / Answer
Do a force balance on piston determines equilibrium float pressure. Piston: mp = (Ap)(l)() Pext = Patm + mpg Ap = 101 kPa + (Ap)(0.1 m)(8000 kg/m3)(9.81 m/s2) (Ap)(1000) = 108.8 kPa The pin is released, and since P1 > Pext, the piston moves up. T2 = T0, so if the piston stops, then V2 = V1 × H2/H1 = V1 × 150/100. Using an ideal gas model with T2 = T0 gives P2 = (P1)(V1/V2) = (200)(100/150) = 133 kPa > Pext P2 = 133 kPa Therefore, the piston is at the stops for the ideal gas model. Now for the tabulated solution: To find P2, we must do some extrapolation because Table B.3.2 does not list 200 kPa. We first interpolate to find the specific volume at T1 = 290 K = 16.85 C at 400 kPa and 800 kPa, and then extrapolate using those points to find the specific volume at 200 kPa, v1. Since v1 = V1 mCO2 and the area of the piston is Ap = (100 mm)2 4 = 0.00785 m2, the mass of carbon dioxide in the cylinder is mCO2 = V1 v1 = (Ap)(100 mm) 0.1682 m3/kg = (0.00785 m2)(0.1 m) 0.1682 m3/kg = 0.000785 m3 0.1682 m3/kg = 0.00467 kg Since the cylinder is closed, the mass of the carbon dioxide does not change, but the volume does, so we will assume the piston is at the stops and V2 = 0.00118 m3, so v2 = V2 mCO2 = 0.2523 m3/kg. Now we have two state variables (T2 and v2), so let’s go back to Table B.3.2 and find a third state variable, P2. Interpolating between 400 kPa and 800 kPa gives P2 = 291 kPa, which means our original assumption that the piston is against the stops was incorrect. A similar iterative approach could be used to find P2 in the tables. The piston is not against the stops. This problem was difficult, so it was only worth two points. You received two points for a reasonable attempt with some calculations, one point for just writing something, and no points for not attempting it. +2 indicates you were awarded two points for part (b), and they were not bonus points