Calculate the work done during an isothermal reversible expansionof a van der wa

Question

Calculate the work done during an isothermal reversible expansionof a van der waals gas. Account physically for the way inwhich th ecoefficients a and b appear in the finalexpression. Plot on the same graph the indicator diagrams forthe isothermal reversible expansion of a) a perfect gas, b) a vander waals gas in which a=0 and b=5.11x10-2dm3mol-1, and c) a=4.2dm6atmmol-2 and b=0. The values selected exaggerate the imperfections but give rise to significanteffects on the indicator diagrams. Take vi = 1.0dm3, n=1.0 mol, and T=298 K.

My teacher gave me the answer from his solutions but I don't get the equations for part B and C. Someone please help.

w0 = -nRTln(Vf/Vi) = ( - 1.0 mol-1) times (8.314 J K-1 mol-1) times (298 K) times ln(2.0 dm3/1.0 dm3) w0 = -1.72 times 103 J = -1.7 kJ w = w0 - (1.0 mol)2 times [0 - (5.11 times 10-2 dm3 mol-1) times (8.314 J K-1 mol-1) times (298 K)] times (1/2.0 dm3 - 1/1.0 dm3) = (-1.72 times 103 J) - (63 J) = -1.78 times 103 J = -1.8 kJ w = w0 - (1.0 mol)2 times (4.2 dm6 atm mol-2) times (1/2.0 dm3 - 1/1.0 dm3) w = w0 + 2.1 dn3 atm = (-1.72 times 103J) + (2.1 dm3 atm) times (1m/10dm)3 times (1.01 times 105Pa/1atm) = (-1.72 times 103 J) + (0.21 times 103 J) = -1.5 KJ Schematically, the indicator diagrams for the eases (a), (b). and (e) would appear as in Figure 2.2. For case (b) the pressure is always greater than the perfect gas pressure and for case (c) always less. Therefore. v2 v1 pdV(c) w(c). Figure 2.2

Explanation / Answer

the gas equation for the van der waal gas is

(P- an^2/V^2)(V-nb)= nRT

for isothermal expansion dW=-PdV

from gas equation, P = nRT/(V-nb) + an^2/V^2

substituting this in work equation and integrating it from initial vol to final vol. u get

W = RT ln( V-nb) + an^2/V

u can use this to find the work .

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