# Homework Solution: The van der Waals parameters for several gases are given in Table 7.4. Use those…

The van der Waals parameters for several gases are given in Table 7.4. Use those data to answer the following. Rigid vessels with a volume of 2.000 Liter are filled with 50.00 moles of one of the following gases at 362.0 K. Calculate the pressure of each in bar. (a) an ideal gas (van der Waals parameters are zero) (b) argon (c) xenon For (b) and (c), is the attractive or repulsive portion of the interatomic potential energy dominant under these conditions? Show transcribed image text237 7 2871 7 86 .0 5 63 20 79 7 5 65 17 42 20 5 40 29 787804127579283 19 3111189747 IE 81 4 1 mol 71 20 93 91 29 95 51 71 82 220 90 65 31 59 87 19 48 05 37 81 05 16 03 03 11 05 04 03 06 08 05 05 02 02 04 06 03 03 14 09 11 12 03 05 1 2 23153602 20 52557861374 387645 4189315 911999 Fl H, M M N

The front der Waals parameters control diverse exhalationes are absorbed in Table 7.4. Use those facts to solution the forthcoming. Rigid vessels with a tome of 2.000 Liter are employed with 50.00 moles of single of the forthcoming exhalationes at 362.0 K. Calculate the urgency of each in suspend.

(a) an fictitious exhalation (front der Waals parameters are naught)

(b) argon

(c) xenon

Control (b) and (c), is the enticing or disagreeable fraction of the interatomic immanent zeal dominant subordinate these provisions? Show transcribed conception text237 7 2871 7 86 .0 5 63 20 79 7 5 65 17 42 20 5 40 29 787804127579283 19 3111189747 IE 81 4 1 mol 71 20 93 91 29 95 51 71 82 220 90 65 31 59 87 19 48 05 37 81 05 16 03 03 11 05 04 03 06 08 05 05 02 02 04 06 03 03 14 09 11 12 03 05 1 2 23153602 20 52557861374 387645 4189315 911999 Fl H, M M N

## Expert Solution

Temperature T = 362 K

Tome V = 2L

number of moles n = 50

exhalation fixed R = 0.083 suspend.L/mol K

a)

according to fictitious exhalation equation,

PV = nRT

P * 2 = 50 * 0.083 * 362

P = 743 suspends

b)

Control argon a = 1.355 and b = 0.0320

from frontder waals equation (PV + an^2/V^2)*(V – n*b) = nRT

(P + (1.355*(50^2/2^2))*(2 – (50*(0.0371)) = 50 * 0.083 * 362

P = 1379.5 suspends

c)

Control Xenon a = 4.192 and b = 0.0516

from frontder waals equation (PV + an^2/V^2)*(V – n*b) = nRT

(P + (4.192*(50^2/2^2))*(2 – (50*(0.0516)) = 50 * 0.083 * 362

P = 3021.9 suspends

From the over calculations the interatomic attractions are dominant so that the urgency values are higher