For an ideal gas undergoing adiabatic compression, the entropy change is:
For a reversible adiabatic process, dq = 0, therefore ΔS = ∫dq_rev/T = 0. Entropy remains constant during reversible adiabatic processes.
In a constant pressure process, the heat absorbed by a system equals:
At constant pressure, q_p = ΔH (change in enthalpy). This is the definition of enthalpy and is a key relationship in engineering thermodynamics.
For a binary ideal solution, the total vapor pressure at constant T is given by:
Raoult's law states P_i = P_i°x_i for ideal solutions. Total pressure P = P₁°x₁ + P₂°x₂. This assumes ideal mixing behavior.
During a constant volume process, the first law of thermodynamics simplifies to:
At constant volume, ΔV = 0, so w = -P∫dV = 0. Therefore, ΔU = q + w = q + 0 = q. All heat goes into internal energy change.
A gas undergoes an isothermal expansion from 2 L to 5 L at 298 K. If the process is reversible, what is the sign of entropy change for an ideal gas?
For isothermal expansion of an ideal gas, ΔS = nR ln(V_f/V_i) = nR ln(25) > 0. Volume increases, so entropy increases.
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In the van der Waals equation, the term 'a' represents:
In (P + a/V²)(V - b) = RT, 'a' corrects for intermolecular forces reducing pressure, while 'b' corrects for molecular volume.
What is the Clausius-Clapeyron equation used for?
Clausius-Clapeyron equation: ln(P₂/P₁) = -(ΔH_vap/R)(1/T₂ - 1/T₁) describes phase equilibrium.
For a system at equilibrium, the chemical potential of a substance in different phases must be:
At phase equilibrium, μ_liquid = μ_vapor = μ_solid. This equality determines equilibrium conditions.
What does the Gibbs free energy criterion ΔG < 0 indicate for a process?
ΔG < 0 indicates spontaneous, irreversible process under constant T and P conditions.
A system absorbs 500 J of heat and does 300 J of work. The change in internal energy is:
First law: ΔU = Q - W = 500 - 300 = 200 J (using convention W = work by system).
