We have a cylinder with 2.0 mol N_2, along with pentane (C5) and hexane (C6). Th

Question

We have a cylinder with 2.0 mol N_2, along with pentane (C5) and hexane (C6). The cylinder has a moveable piston in it, which has a weight to keep P = 5.5 bar. At the initial temperature T_1 = 295 K, the chamber has a volume V_1. Then the temperature is increased to T_2 = 350 K, giving a new volume V_2. The liquids have a negligible volume, but both at 295 K and 350 K, there remains at least 1 drop of liquid. The Antoine equations for the vapor pressure of pentane and hexane are given near the last pages of this exam. What is the change in volume in the cylinder? V_2 - V_1 = m^3

Explanation / Answer

Solution:

By using antoine equation let we find pressure change at 350K,

log10(P) = A (B / (T + C))
P = vapor pressure (bar)
T = temperature (K)

Pentane   -----> A= 13.8193, B=2696.04, C = 224.317

Hexane    -----> A= 13.7667, B=2451.88, C=232.014

P = Ppentane + Phexane = (10(13.8193 -(2696.04/350+224.317)) +(10(13.7667-(2451.88/350+232.014))

   = 12.8 bar

V1= nRT1 / P1 = (8.314*295)/5.5 = 445.9327 cu.m

V2 = nRT2 / P2 = (2*8.314*350)/12.8 = 454.671 cu.m

V2-V1 = 454.671 - 445.9327 = 8.7383 cu.m

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