Termodinamica_quimica

expansión isoteŕmica\nt1\nQ=0\ncompresión adiab́atica\nT2\nexpansión adiab́atica\ncompresión isoteŕmica\nt1 q = w = RT ∫dV/V\nq = w = ∫pdV ∫ pdV CH3OH(64°C e 1 atm) → CH3OH(64°C e 0.5 atm)\nΔH = m · ΔHsup ΔH a ΔH vap\nΔH a = 32 g/mol · 260 cal/g = 8320 cal/mol\nΔG a = 0\nCH3OH(g)(64°C e 1 atm) → CH3OH(g)(64°C e 0.5 atm)\ndG = -S · dT + V · dp\nΔH = 0\n\ndG = V r · dp\n\ndG r − dG = R · T · ln P1/P1\n\ndG r = 1.987 cal·mol⁻¹·K⁻¹ · 337 K · ln 0.5 atm/1 atm\nΔG = -464 cal/mol ΔH T = ΔH a + ΔH = 8.320 + 0 = 8.320 cal/mol\nΔG T = ΔG a + ΔG = 0 + (-464) = -464 cal/mol