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One mole of a gas is allowed to expand isothermally and reversibly from a volume of 1dm-3 to 50dm-3 at 273k . calculate the work done assuming van DER wall behavior.(given a=6.5atm dm-6mol-2,b=0.0056dm-3mol-2)
Accepted Answer
The work done during the isothermal reversible expansion of a van der Waals gas can be calculated using the following equation:
W = -nRT ln[(V2 - nb) / (V1 - nb)] + a
n^2
[(1 / V2) - (1 / V1)]
Where:
W is the work done
n is the number of moles (1 mol)
R is the ideal gas constant (8.314 J/mol
K)
T is the temperature (273 K)
V1 is the initial volume (1 dm^3 = 0.001 m^3)
V2 is the final volume (50 dm^3 = 0.05 m^3)
a is the van der Waals constant for attraction (6.5 atm dm^6/mol^2 = 657.8 J
m^3/mol^2)
b is the van der Waals constant for volume (0.0056 dm^3/mol = 5.6 x 10^-6 m^3/mol)
Plug in the values and calculate the work done. Remember to convert units to be consistent (e.g., atm to Pa, dm^3 to m^3).
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