Derive the general expression for the work per kilomole of a van der waals gas in expanding reversibly and at a constant temperature, T, from a specific volume V1 to a specific volume V2.
To derive the general expression for the work per kilomole of a van der Waals gas during reversible expansion at a constant temperature, we need to use the first law of thermodynamics, which states that the work done by a system is equal to the change in its internal energy (ΔU) plus the heat transfer (q).
In this case, since the expansion is reversible and at constant temperature (T), there is no change in internal energy (ΔU = 0). Therefore, the work done is equal to the heat transfer (q) from the system.
To find the expression for work per kilomole, we can use the ideal gas equation and the van der Waals equation of state. The ideal gas equation is given by:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
The van der Waals equation of state accounts for the intermolecular forces present in real gases and is given by:
[P + a(n/V)^2][(V - nb)] = nRT
where "a" and "b" are constants specific to each gas, representing the attractive and the repulsive forces between the gas molecules.
Now, we can differentiate the van der Waals equation of state with respect to volume V and rearrange to solve for pressure P:
dP = [an^2 / V^3 - 2abn / V^2 + nRT / V^2]dV
Next, we can substitute the ideal gas equation (PV = nRT) into the last term of the derivative, leading to:
dP = [an^2 / V^3 - 2abn / V^2 + P / V^2]dV
Since we are interested in determining the work done during the reversible expansion, we can integrate both sides of the equation from V1 to V2:
∫dP = ∫[an^2 / V^3 - 2abn / V^2 + P / V^2]dV
Upon integration, the left-hand side gives the pressure difference:
P2 - P1 = ∫ [an^2 / V^3 - 2abn / V^2 + P / V^2]dV
Now, we can rearrange the equation and solve for P, giving:
P = [(an^2 / V^3 - 2abn / V^2) + P / V^2]
Substituting back the value of P into the equation and simplifying further, we get:
∫dV / V^2 = ∫[an^2 / V^3 - 2abn / V^2 + P / V^2]dV / P
The left-hand side of the equation integrates to:
-(1 / V) = -(1 / V2) + (1 / V1)
Now, we can solve for the integral of the right-hand side by breaking it down into three separate integrals:
∫[an^2 / V^3 - 2abn / V^2 + P / V^2]dV / P = ∫an^2 / V^3 dV / P - ∫2abn / V^2 dV / P + ∫P / V^2 dV / P
Upon integration, the first integral simplifies to:
(an^2 / P) * [1 / V2 - 1 / V1]
The second integral simplifies to:
(2abn / P) * [1 / V2 - 1 / V1]
And the third integral gives:
(P / P) * [1 / V2 - 1 / V1]
Simplifying further, we can substitute these expressions back into the equation:
-(1 / V) = (an^2 / P) * [1 / V2 - 1 / V1] - (2abn / P) * [1 / V2 - 1 / V1] + [1 / V2 - 1 / V1]
Multiplying through by -P, we obtain:
(1 / V) = (an^2 / P) * [1 / V2 - 1 / V1] - (2abn / P) * [1 / V2 - 1 / V1] + [1 / V2 - 1 / V1]
Now, we can solve this equation for the work done per kilomole of the van der Waals gas during reversible expansion:
W = -PΔV
Substituting the expression for P from the previous equation, we get:
W = -[(an^2 / V2) * (1 / V2 - 1 / V1) - (2abn / V2) * (1 / V2 - 1 / V1) + (1 / V2 - 1 / V1)] * (V2 - V1)
Simplifying further:
W = -[an^2 / V2^2 - 2abn / V2 - 1 / V2 + an^2 / V2V1 - 2abn / V1 - 1 / V1 + 1 / V2 - 1 / V1] * (V2 - V1)
Finally, if we multiply by the number of moles (n), we obtain the general expression for the work per kilomole of a van der Waals gas in reversible expansion at constant temperature:
W = -n[an^2 / V2^2 - 2abn / V2 - 1 / V2 + an^2 / V2V1 - 2abn / V1 - 1 / V1 + 1 / V2 - 1 / V1] * (V2 - V1)
This is the desired general expression for the work per kilomole of a van der Waals gas in reversible expansion.