![]() ![]() At this point, you divide ΔH 85degC by the temperature (358.15 K) to get the entropy. So, you can see that your method is the exact same process as method 1, where you add the change in enthalpy from 85 degC up to 100 degC, add the ΔH vap at the normal temperature, and subtract the enthalpy of cooling the vapor back down to 85 degC. ΔH 85degC = ΔH 100degC + ΔT (C p liquid - C p vapor) Here, we add in ΔH vap at the normal temperature, 109 J/K/mol * 373.15 KĬombining this and equation (5), we have Kirchhoff's Law, In step 2 from the initial method, we added ΔS vap at the normal temperature. ( 5) C p liquidΔT - C p vaporΔT = ΔT(C p liquid - C p vapor) However, we are also going to subtract the change in enthalpy of the vapor going from 100 degC to 85 degC (step 3 in method 1). So, following method 1, but using enthalpy, we take at 358.15 K (85 degC) ![]() Our C is the heat capacity at constant pressure. In this case, we are talking about the molar enthalpy, so the mass in (1) is simply 1 mole, and it is divided through to the left side, and we end up with To prove this, you may remember from Gen. Your second method is the exact same process, except you deal in enthalpy. This is your first method, and it is the correct answer. ![]() subtract the change in entropy as the vapor is cooled from 100 deg C to 85 deg C. sum the change in entropy from 85 degC to 100 deg Cģ. If someone could tell me what I did wrong, I would really appreciate it! I thought to find the :delta: H vaporization at 100☌ by doing 373.15K *109.0 J/K*mol = 40673.35 J/mol. However, I initially did it a different way and wanted to know what was wrong with what I did: The answer obtained from this method is 111 J/K*mol. I realize that it is possible to do this problem by finding the entropy change to heat liquid water from 85☌ to 100☌, finding the entropy change in vaporizing this liquid water (which would just be 109.0J/K*mol), and finding the entropy change to lower the temperature of the water vapor back to 85☌. Entropy of vaporization is an increase in entropy upon vaporization of a liquid.Calculate the standard entropy of vaporization of water at 85☌, given that its standard entropy of vaporization at 100.☌ is 109.0 J/K*mol and the molar heat capacities at constant pressure of liquid water and water vapor are 75.3 J/K*mol and 33.6 J/K*mol, respectively, in this range. Note: Enthalpy of vaporization is the amount of energy that must be added to the liquid substance, to transform a quantity of that substance into gas. This maximum can be attained only in a completely reversible process. 1 atm.In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of irreversible work that may be performed by a thermodynamic system at a constant temperature and pressure.The Gibbs free energy is the maximum amount of non-expansion work that can be extracted from a thermodynamic closed system. ![]() The change in free energy occurs when a compound is formed from its elements in their most thermodynamically stable state at standard state conditions i.e. ∴ Boiling point of the liquid at one atmospheric pressure is 400k. $\Delta H$ = enthalpy of vaporization = 30000 $Jmo$ It is given that Enthalpy of vaporization is 30 Kilojoules per mole. We know that at equilibrium, Gibbs energy change is zero. vapor is in equilibrium at one atmospheric pressure. Hint: the fact that at boiling point of the liquid, liquid & its vapor are in equilibrium i.e. ![]()
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