a) A 0.250m3 tank contains 200mol of ammonia at 20C. Heat is added until tempera
ID: 538592 • Letter: A
Question
a) A 0.250m3 tank contains 200mol of ammonia at 20C. Heat is added until temperature rises to 40C. Determine the amount of heat.
b) One mol of ammonia at 8.6 bar, 100C, is cooled under constant pressure by removing 25 kJ=mol of heat. Determine the nal temperature. If the system is a vapor/liquid mixture, report the mol fraction of the liquid.
c) One mol of ammonia at 0C, 8.6 bar, is mixed adiabatically at constant pressure with 1 mol of ammonia at 100C, 8.6 bar. Determine the nal temperature. If the system consists of a vapor-liquid mixture, report the mol fraction of the liquid.
*** BELOW IS SOLUTION TO a) AND b) BUT WHAT ARE THE IN BETWEEN STEPS??
The following data are available for ammonia: Saturated ammonia Vy (rm3/mol) UL (kJ/mol) T (°C) P (bar) (m3/mol) Uv (k//mol) 25.225 25.357 25.467 25.552 25.610 25.637 25.627 25.573 25.464 25.284 25.000 /11 (k//mol) 5.844 6.637 7.441 8.258 9.091 9.942 10.816 11.720 12.662 13.654 14.717 //v (kJ/mol) 27.340 27.508 27.645 27.748 27.811 27.831 27.801 27.710 27.546 27.288 26.904 4.3 2.667 x 10 4.927 x 103 5.833 106.2 2.726 x 105 3.499 x 106.620 7.417 30 1.7 2.861 x 105 881 x 10-3 8.225 0 15.6 2.939 x 10-.415 x 103 9.045 9.880 10.735 4 11.613 0 12.522 13.474 14.484 20 8.6 2.791 x 10 2.541 x 10-3 50 60 26.2 3.123 × 1058.310×104 70 80 41.4 3.368 x 105 5.026 104 90 100 62.6 3.730 x 10 3.035 x 104 20.3 3.026 x 10.079 x 103 33.1 3.236 x 106.449 x 10 51.2 3.528 x 105 3.917x 104 The constant-pressure heat capacitics of the vapor and liquid are =80.8 J/mol K; C: = 51.6J/mol K.Explanation / Answer
a.
In the sterp one to determine amount of liquid and vapor
we use formula of lever rule
X = (Vvap-Vtank)/Vvap- Vliq)
for 20 C we get x=.51371
similarly for 40 c we get x= .11981
to find change in Q we simply subtract Q20 from Q40
b.
In th first step H for given conditions is calculated because it is not given directly in the table so we use above formula
we also know 25 KJ energy was emoved
this means diffeence in H = 25 KJ
from this we get final H = 6.773 KJ
but this is less than enthalpy of saturated liquid at that state which is 7.441 KJ/mol
So the temperature can not betaken from the table directly but calculated using above formula.
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