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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)work it all till the end and give answer
Accepted Answer
The work done during an 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/V2 - n^2/V1)
Where:
W is the work done
n is the number of moles (1 mole in this case)
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)
V2 is the final volume (50 dm^3)
a and b are the Van der Waals constants (6.5 atm dm^6/mol^2 and 0.0056 dm^3/mol^2, respectively)
Substituting the given values into the equation, we get:
W = -(1 mol)(8.314 J/mol·K)(273 K) ln((50 dm^3 - (1 mol)(0.0056 dm^3/mol^2)) / (1 dm^3 - (1 mol)(0.0056 dm^3/mol^2))) - (6.5 atm dm^6/mol^2)(1 mol)^2/(50 dm^3 - (1 mol)(0.0056 dm^3/mol^2)) - (6.5 atm dm^6/mol^2)(1 mol)^2/(1 dm^3 - (1 mol)(0.0056 dm^3/mol^2))
Solving the equation gives us:
W = -7.34 x 10^3 J
Therefore, the work done during the isothermal reversible expansion of the Van der Waals gas is -7.34 x 10^3 J.
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