Question

Equal molar amounts of neon were collected over water in two identical 6.0-L flasks, one being at 30 degrees Celcius and the other being at 20 degrees Celcius, respectively. The difference of respective total pressures in two flasks was 845.8 torr. Use the provided table of vapor pressure versus temperature (picture) to calculate the number of moles of neon collected in each flask.

uuulements Vapor Pressure of Water at Selected Temperatures (*) Temperature, T°C Vapor pressure, I Pvapor Torr I Temperature, T, °C Vapor pressure, P vapor: Torr 15 12.8 23 21.0 16 13.6 24 22.4 17 14.5 25 23.8 18 15.5 26 25.2 19 27 26.7 20 16.5 17.5 18.6 28 28.3 29 21 22 30.0 31.8 19.8 30

Answer 1

Flask A :

T = 30℃ = 303.15K , V = 6.0L , P_Ne = (P_total - 31.8)Torr

n_A = x

Flask B:

T = 20℃ = 293.15K , V = 6.0L , P_Ne = (P_total - 17.5)Torr

n_B = y

P_total (A) - P_total(B) = 845.8torr

P_Ne,A + 31.8 - P_Ne,B - 17.5 = 845.8

P_Ne,A - P_Ne,B = 831.5torr

(n_Ne,A RT_A - n_Ne,B RT_B )/V = 831.5 torr

303.15*x - 293.15*y = (831.5/760)×6.0/0.0821 = 79.957

x - 293.15/303.15*y = 79.957/303.15

x - 0.967*y = 0.2638

Number of moles of water in flask A = PV/RT = (31.8/760)×6.0/(0.0821×303.15) = 0.01009mol

Number of moles of water in flask B = (17.5/760)×6.0/(0.0821×293.15) = 0.00574mol

x + 0.01009 = y + 0.00574

x = y - 0.004347

y - 0.004347 -0.967*y = 0.2638

y = 0.2681/0.03299

y = 8.12747mol

x = 8.12747 - 0.004347 = 8.12313 mol

Number of moles of Ne in flask at 30℃ = 8.1231mol  

Number of moles of Ne in flask at 20℃ = 8.1275mol