Question

Consider 5.000 mol of neon in a 208.7 cm^3 piston at 120.0 K. If the gas undergoes isometric cooling to one-half of its initial temperature and then undergoes isothermal reversible expansion to four times of its initial volume, (a) compute the work done assuming the gas behaves ideally. Next, (b) compute the work done for the entire process assuming the gas behaves as a van der Waal's gas (a = 0.2050 atmL^2/mol^2, b = 1.670 x 10^-2 L/mol). (c) Compare and explain your results.

Answer #1

a.] work done in isothermal reversible expansion is given as:

w = -nRT ln (Vf/Vi) where Vf = final volume and Vi = initial volume

T = 120/2 = 60 K

Vi = 208.7 cm^3

Vf = 4 X 208.7 cm^3 = 834.8 cm^3

Put in the values in the above equation:

w = -5 mol X 8.314 J / K.mol X 60 K X ln (208.7 cm^3 / 834.8 cm^3)

w = - 2494.2 J X (-1.386)

**w = 3457.695 J**

b.] w = nRT ln [(Vf - nb) / (Vi - nb)] + n^{2}a [(1/Vf)
- (1/Vi)]

w = 5 mol X 0.0821 L atm/K.mol X 60 K X ln[(0.8348 L - 0.0835 L)
/ (0.2087 L - 0.0835 L)] + 5.125 atm.L^{2} [(1.198 L -
4.792 L)]

w = 24.63 L.atm X ln (0.7513 / 0.1252) + 5.125 atm.L^{2}
(-3.594 L^{-1})

w = 44.134 L.atm - 18.42 atm.L

**w = 25.714 L.atm**

Please note that for part B, R used is 0.0821 L.atm / mol.K , all volumes in cm^3 have been converted to L (1 cm^3 = 0.001 L).

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