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Calculate the pressure exerted by benzene for a molar volume of 2.10 L at 610. K using the Redlich-Kwong equation of state: P==RTVm−b−aT√1Vm(Vm+b)nRTV−nb−n2aT√1V(V+nb) The Redlich-Kwong parameters a and b for benzene are 452.0 bar⋅dm6⋅mol−2⋅K1/2 and 0.08271 dm3⋅mol−1, respectively. Express your answer with the appropriate units.

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ajeigbeibraheem

Answer:

[tex]P=21.148 bar[/tex]

Explanation:

Given parameters are ;

Molar Volume (V_m) = 2.10 L

Temperature (T) = 610 K

a = 452.0 bar dm⁶mol⁻²/[tex]K^{(1/2)[/tex]

b = 0.08271 dm³ mol⁻¹

Our given formula to use is Redlich-Kwong equation of state, which is given as:

[tex]P = \frac{RT}{V_m-b}-\frac{a}{\sqrt{T} } *\frac{1}{V_m(V_m+b)} =\frac{nRT}{V-nb}-\frac{n^2a}{\sqrt{T} }\frac{1}{V(V+nb)}[/tex]

[tex]P = \frac{RT}{V_m-b}-\frac{a}{\sqrt{T} } *\frac{1}{V_m(V_m+b)}[/tex]

[tex]P = \frac{8.314*10^{-2} bardm^3mol^{-1}k^{-1}*610K}{2.10dm^3mol^{-1}-0.08271dm^3mol^{-1}}-\frac{452bardm^6mol^{-2}K^{1/2}}{\sqrt{610K} } *\frac{1}{2.10dm^3mol^{-1}(2.10dm^3mol^{-1}+0.08271dm^3mol^-}[/tex]

[tex]P = \frac{50.72}{2.01729}- 18.30095*0.2182[/tex]

[tex]P= 25.14-3.992[/tex]

[tex]P=21.148 bar[/tex]