An ideal gas at 0 °C and 10 atm has the volume of 10 dm3. Calculate the final volume and work done u — Physical Chemistry — Thermodynamics Chemistry Question
Work in Thermodynamics
An ideal gas at 0 °C and 10 atm has the volume of 10 dm3. Calculate the final volume and work done under the following three sets of conditions if the final pressure is 1 atm.
Isothermal reversible expansion.
Model Answer
We have 100 / 22.41 = 4.461 moles, and the final volume is:
V2 = p1 * V1 / p2 = 10 * 10 / 1 = 100 dm3
The work done by the gas is:
w = -q = -n * R * T * ln(V2 / V1) = -4.461 * 8.314 * 273.2 * ln(10) = -23335 J
Adiabatic reversible expansion.
Model Answer
Notice that:
γ = Cp / Cv = (Cv + R) / Cv = (1.5 * R + R) / (1.5 * R) = 5/3
Thus, the final volume is:
V2 = V1 * (p1 / p2)^(1/γ) = 10 * (10 / 1)^(3/5) = 39.8 dm3
And the final temperature is obtained from:
T2 = p2 * V2 / (n * R) = 1 * 39.81 / (4.461 * 0.08205) = 108.8 K
For adiabatic processes:
q = 0 and ΔE = q + w = w
i.e.,
w = ΔE = n * Cv * ΔT = 4.461 * 1.5 * 8.314 * (108.8 - 273.2) = -9141 J
Irreversible adiabatic expansion is carried out as follows: Assume that the pressure is suddenly decreased to 1 atm and the gas expands adiabatically at a constant pressure.
[Note that the molar heat capacity at constant volume is given by the relation: Cv = 3/2 R, where R is the gas constant.]
Model Answer
Since q = 0, we have:
ΔE = w = n * Cv * (T2 - T1)
w = -p2 * (V2 - V1)
And:
3/2 * n * R * (T2 - 273.2) = -p2 * (n * R * T2 / p2 - n * R * T1 / p1)
This simplifies to:
3/2 * (T2 - 273.2) = -(T2 - 273.2 * 1 / 10)
It follows that:
T2 = 174.8 K
And:
w = ΔE = n * Cv * ΔT = 1.5 * n * R * (174.8 - 273.2) = -5474 J