Problem Set #6: Van der Waals Gas Calculations

University:
The Pennsylvania State University
Course:
CHEM 450 | Thermodynamics
Academic year:

2024
Views:

332

Pages:

3
Author:

Alexis H.

Problem Set #6 Assigned: January 22, 2024 Due: January 24, 2024 at 2:30 pm Upload your answers to the Canvas website. PDF files are preferred, but Microsoft Word, or image files will be accepted. Please check to make sure that your submission is legible. 1. Calculate the pressure exerted by 1.0 mol C2H6 behaving as a van der Waals gas under the following conditions: (a) 273.15 K and 22.414 dm3 and (b) 1000 K and 100 cm3. (note: a = 5.507 atm L2/mol2 and b = 0.0651 L/mol) 2. Express the van der Waals coefficient a = 0.751 atm L2/mol2 and b = 0.0226 L/mol in SI based units (kg, m, s, and mol). 3. A gas at 250 K and 15 atm has a molar volume 12% smaller than that calculated from the perfect gas law. (a) Calculate the compression factor, Z, under these conditions. (b) Find Vm for the gas. (c) Do the gas molecules experience net attractions or repulsions under these conditions. 4. Consider a 1.000 m3 volume of N2 at 500 K. The mass of the gas is 92.4 kg. Use the van der Waals equation of state to determine the pressure of the gas. 2 (note: a = 1.352 /mol2 and b = 0.0387 L/mol) 5. Suppose that 10.0 mol C2H6 gas is confined inside a 4.860 L container at 27 °C. (a) Predict the pressure exerted by the gas if it behaves as a perfect gas and as a van der Waals gas. (b) Find the compression factor. 2 /mol2 and b = 0.0651 L/mol) 1 p V2 1m01 0.0821 69th 273.15k 22.414dm 0.065 my Imo 1m01 5.507 1m01 0.0821 p 22.414dm 100dm 097 Imo 1000k 1m01 0.065 my 5.507 100dm P 0 821 atm III 2 P MRT Pt Pt V nb 4 NRT v nb V nb P A D V p a Pa m6 mg a 0.751atm dm 2 mom a 0.076Pa m6 no D m Mot t D 0.0226dm molt 0 2.26 10 5 m Mol I 3 P 15atmT 250kR 0.0821ho9thh a V Vm V I Vm v z VM b Vm V 0.88 I 1 0.12 0.88 4 mm 0.9924 atm P b n a TRI P a 88 010 0.88 Vm 1.21L 0.88 250 Ifhm fm 0821 0.88 mm c net attractions 4 92.4kg 92400g 92400g 288134g P P a III 3298.42mol 0.08206 tho l dm P 5 3298.42mol 1.352 500k 1dm 3298.42mol 0.03 Lymo 14.7 10 atm fp P a 10mol 0.082158th 300.15k 4.860dm P 50 7 atm P p II 9 tl 10m01 0.0821ho9 5 507 10mol 4.860dm 4.860dm 300.15k 10mo 0.06g P 35.23atm b Z z 1 35.23atm 4860dm 10m01 0.08206 300.15k Z 0.695 9mmol 3298.42mol 2