Question

A 107 g sample of CH3CH2Br (Bromoethane) gas has a volume of 4.375 L at a temperature of 36°C.
i. Calculate the pressure using the Ideal Gas Law.
ii. Calculate the corrected pressure using the Van der Waals Equation.
atm×L2 L a=11.723 mol2 b=0.08406 mol

Answer 1

Molar mass of Bromoethane is 109 gms.

Thus, number of moles = $n=107/109$

Given : $V=4.375 L ; T=36°C=273+36=309K$

We know Gas constant = $R=0.0821 Latm mol^{-1}K^{-1}$

i. Ideal gas law : $PV=nRT \implies P=nRT/V$

$P=107*0.0821*309/(109*4.375)=5.692 atm$ ---(Answer)

ii. Van der Waals Equation : $(P+an^2/V^2)(V-nb)=nRT$

Given : $a=11.723atmL^2 mol^{-2}; b=0.08406L mol^{-1}$

Thus, $(P+(11.723*107^2/(109^2*4.375)))(4.375-$$(107*0.08406/109)=107*0.0821*309/109$

$\implies (P+2.582)(4.375-0.0825)=24.9$

$\implies P=(24.9/4.292)-2.582=3.22 atm$ ----(Answer)