ThermodynamicsHard
Thermodynamic cycle (work and entropy)
n = 2 mol of an ideal gas at T₁ = 300 K perform a reversible cycle: isothermal compression from V₁ = 12 L to V₂ = 3 L, adiabatic expansion back to V₁, isochoric heating up to T₁. Find the work of the isotherm and the total entropy change over the cycle.
Given data
n = 2 molT₁ = 300 KV₁ = 12 L, V₂ = 3 LR = 8.314 J/mol·K- In a reversible isothermal transformation the temperature is constant, so the work is the integral of p dV with p = nRT/V: L₁₂ = nRT₁·ln(V₂/V₁). Here it is a compression (V₂ < V₁), so the logarithm is negative: the work done by the gas is negative (the surroundings do work on the gas).
- Entropy is a state function: it depends only on the state of the system, not on the path. Over a cycle the system returns to its initial state, so its total entropy change is necessarily zero, whatever the individual transformations. What is ΔS over the cycle?
Full worked solution
- In a reversible isothermal transformation the temperature is constant, so the work is the integral of p dV with p = nRT/V: L₁₂ = nRT₁·ln(V₂/V₁). Here it is a compression (V₂ < V₁), so the logarithm is negative: the work done by the gas is negative (the surroundings do work on the gas).(compression → negative work).
- Entropy is a state function: it depends only on the state of the system, not on the path. Over a cycle the system returns to its initial state, so its total entropy change is necessarily zero, whatever the individual transformations. What is ΔS over the cycle?: entropy is a state function, so over a reversible cycle the total change is zero.
Result: kJ,