One mole of CO2(g) is expanding isothermally and reversibly from a volume of 0.100 dm3 to a volume of 100 dm3. Calculate the work in "kJ" of the expansion process if: 1/ the gas is considered real gas. (Hint: use the Van Der Waals equation for real gases to express the pressure in the work equation). 2/ The gas is considered perfect gas. Given: The constants for CO2(g) are: a= 3.59 atm L2/mol2, b= 0.0427 L/mol Answer: a) the expansion work for CO2 as real gas is W = kJ. b) the expansion work for CO2 as perfect gas is W = kJ.

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AbsorbingMan AbsorbingMan · Answer 1 · 2021-04-13T08:56:13+02:00

Solution :

For the real gas following Vander Waals equation, reversible isothermal work done is given by :

[tex]$W= \int P \ dV$[/tex]

[tex]$W= \int \left(\frac{nR}{V-nb}-\frac{an^2}{V^2}\right) dV$[/tex]

[tex]$W_{real}=-nRT \ln \left[\frac{V_2-b}{V_1-b}\right]-an^2\left[\frac{1}{V_2}-\frac{1}{V_1}\right]$[/tex]

Given : [tex]$V_1=0.1 \ dm^3$[/tex], [tex]$V_2 =100 \ dm^3[/tex]

[tex]$a=359 \ atm \ L^2/ mol^2, \ \ b = 0.0427 \ L/mol$[/tex]

As T is not given, assuming T = 290 K

[tex]$W_{real} = 0.0823 \times 240 \left[ \ln \left(\frac{100-0.042}{0.1-0.0427}\right)\right]-3.59\left[\frac{1}{100}-\frac{1}{0.1}\right]$[/tex]

[tex]$W_{real}= -183.06+35.864$[/tex]

[tex]$W_{real}=-147.196$[/tex]

For the perfect gas,

[tex]$W=nRT \ln\left[\frac{V_2}{V_1}\right] =(1)(0.0823)(290) \ln \left[\frac{100}{0.1}\right]$[/tex]

[tex]$W_{ideal}=169.415$[/tex] kJ