The mathematical equation for an ideal gas undergoing a reversible (i.e., no entropy generation) adiabatic process can be represented by the polytropic process equation where P is pressure, V is volume, and γ is the adiabatic index or heat capacity ratio defined as WebJan 15, 2024 · For changes in which the initial and final pressures are the same, the most convenient pathway to use to calculate the entropy change is an isobaric pathway. In …
Why is the enthalpy change not zero in an adiabatic process?
WebExpert Answer. 10 moles of ideal gas at 1000 K are adiabatically expanded in a turbine from a pressure of 100 atm to 10 atm. Use Cp = 25R,Cv = 23R (a) Calculate the amount of work produced by this reversible adiabatic expansion. (b) Does the entropy of the system change by this reversible adiabatic expansion? Web89.Entropy change depends on (a)heat transfer (b)mass transfer (c)change of temperature (d)thermodynamic state (e)change of pressure and volume. Ans: a 90.For reversible adiabatic process, change in entropy is (a) maximum (b) minimum (c) zero (d) unpredictable (e) negative. Ans: c. 91.Isochoric process is one in which (a)free … snow on gas meters
Magnetoelastic transition and negative thermal expansion of Fe2
WebJun 13, 2024 · w = q = Δ T = Δ E = Δ H = 0. (free expansion, ideal gas) For an adiabatic free expansion, we have d q = 0 and d w = 0, and it follows again that w = q = Δ T = Δ E = Δ H = 0. We see that the isothermal and adiabatic expansions of an ideal gas into a vacuum are equivalent processes. If the expansion is opposed by a non-zero applied … WebEntropy is a measure of the disorder of a system. Entropy also describes how much energy is not available to do work. The more disordered a system and higher the entropy, the less of a system's energy is available to do work. Although all forms of energy can be used to do work, it is not possible to use the entire available energy for work. WebAug 8, 2024 · Correct answer: Entropy. My answer: Compressibility. My Reasoning : Z = P V n R T. In adiabatic T V γ − 1 = constant. Therefore, T ∝ 1 V γ − 1. γ − 1 is greater than zero always. So we can say is volume increases, temp decreases. Therefore Z should increase as its numerator increases and denominator decreases. snow on golf course picture