Hi, here is your assignment :). I will send you both word file and pdf file :). They were both the same :)

Spring 2019 Due Jan 29th

PNGE 332

HOMEWORK #2
(Submit the typed HW in single pdf file through ecampus)
1. Pure Heptane is contained in a laboratory cell in which the volume can be varied. The
cell is maintained at a temperature of 400°F. The volume is changed until a few drops of
liquid Heptane are present. What is the pressure in the cell? The volume is reduced until
the cell is essentially full of liquid Heptane with only a few bubbles of Heptane gas
remaining. What is the pressure? Explain.
Answer:
Hence, we have to do a chart of vapor pressure of normal paraffin's and with the help of
that chart, we can easily find out the vapor pressure i. e 400 psi. Temperature of 4000f
for pure n-butane. Here we have isothermal system and are substance is pure. So we can
expect that during liquid/gas phase (which is substance is in), the pressure will relatively
constant.
pressure at 4000f
400
300

vapor pressure of n-butane

200
100

100

200

300

400

2. The densities of methane liquid and gas in equilibrium along the vapor-pressure line are
given below. Estimate the density of methane at its critical point of -116.7°F

Temperature °F
-253
-235
-217
-199
-181
-163

Density of saturated methane
Liquid
Gas
lb/ft3
lb/ft3
26.17
0.1443
25.25
0.2747
24.28
0.4766
23.18
0.7744
21.89
1.202
20.48
1.810

Answer:
Here we made the tabulate values shown on the table below as we convert the given temperature in
Fahrenheit to Kelvin:
T(F)
-253
-253
-217
-199
-181
-163

T(K)
114.817
124.8 17
134.817
144.817
154.817
164.817

Liquid
26.17
25.25
24.28
23.18
21.89
20.48

Vapor
0.1443
0.2747
0.4766
0.7744
1.202
1.81

Plot:

At the critical point, the two phase's liquid and vapor becomes indistinguishable and the density of
saturated liquid and vapor are the same. From the above plot, at a critical point of (190.53 𝐾/
116.7℉), the density at that critical point is 10.12 lb/ft3.

3. Estimate the critical pressure of a mixture of methane and n-butane if the critical
temperature of the mixture is known to be 200°F. Compare your answer with the critical
pressures of pure methane and pure n-butane (Use Appendix A).
Answer:
The Critical pressure of methane (1):
45.99 𝑏𝑎𝑟
= 46 ∗ 10^5 𝑝𝑎𝑠𝑐𝑎𝑙
The critical pressure of pure hexane (2) :
30.20 𝑏𝑎𝑟
= 30.2 ∗ 10^5 𝑝𝑎𝑠𝑐𝑎𝑙
We have from van der waals law that:
𝑃𝑐𝑉𝑐
𝑅𝑇𝑐
8
=
3
= 0.375
Vc1=
𝑅𝑡𝑐1
0.375 ∗
𝑝𝑐1
366.33
= 8.314 ∗
∗ 105
46
= 2.483 ∗ 10^ − 4 𝑚^3
Vc2=
𝑅𝑡𝑐2
0.375 ∗
𝑝𝑐2
366.33
= 8.314 ∗
∗ 105
30.2
= 3.782 ∗ 10^ − 4 𝑚3
We have from vanderwaals law that,
𝑃𝑐𝑉𝑐
𝑅�...