Scienc

Thanks. Thank so much,

I am a beginner in QE and I have a question. During I was reading about Output file, I saw a part where have it: "The calculation goes on step by step until convergence is reached. Then we show all the energies of all bands (eigenvalues of the Kohn-Shan orbitals) for each k-point, the Fermi energy, the total energy, the contributions of all the terms to the total energy, and the calculation time used by each subroutine of the program." Output file: ... iteration # 9 ecut= 100.00 Ry beta=0.70 Davidson diagonalization with overlap ethr = 1.00E-13, avg # of iterations = 1.0 negative rho (up, down): 1.151E-06 0.000E+00 total cpu time spent up to now is 28.4 secs End of self-consistent calculation k = 0.0000 0.0000 0.0000 ( 3909 PWs) bands (ev): -17.8569 -5.9595 -1.3957 -1.3957 4.4242 9.2594 9.9561 9.9561 k = 0.0000 0.0550 0.0000 ( 3909 PWs) bands (ev): -17.8034 -5.8955 -1.6155 -1.5115 4.4968 9.3266 9.9837 10.1886 ... -10.9411 -10.9411 -8.9101 1.6562 1.6562 12.4084 14.2298 14.2298 the Fermi energy is 1.6562 ev ! total energy = -22.80493136 Ry Harris-Foulkes estimate = -22.80493136 Ry estimated scf accuracy < 1.5E-15 Ry The total energy is the sum of the following terms: one-electron contribution = -44.02527202 Ry hartree contribution = 24.36506286 Ry xc contribution = -6.99363523 Ry ewald contribution = 3.85762303 Ry Dispersion Correction = -0.00870488 Ry smearing contrib. (-TS) = -0.00000512 Ry convergence has been achieved in 9 iterations Writing output data file grafeno.save init_run : 2.26s CPU 2.29s WALL ( 1 calls) electrons : 20.10s CPU 20.24s WALL ( 1 calls) ... Parallel routines fft_scatter : 4.35s CPU 4.44s WALL ( 23020 calls) PWSCF : 28.46s CPU 28.70s WALL This run was terminated on: 11:51: 0 5Sep2017 =------------------------------------------------------------------------------= JOB DONE. =------------------------------------------------------------------------------=

So, is the true value for work +22, because work is done on the system? I used the 𐤃U = q+w. because I studied in a chemical class. However, I considered the contribution to work as negative. The book on the other hand, used a positive signal for work. The images is from the book.

I have a question about the resolution of this exercise: I tried to solve this problem with the following steps: For the 1º law of thermodynamics, 𐤃U = q+w. As q = 𐤃H at constant pressure, I assumed that in this process q = - 15 kJ. As a result, 𐤃U = -15 - 22 (because was an expansion) and for this reason, 𐤃U = -37 kJ. However, the book has this resolution I understood why the book used this method, but I did not understand why my method is wrong.

Why in an irreversible process, is the pressure constant? I understood why I should use Wirrev = -PΔV, but I did not understand why the pressure is constant in this process.

Why does CH3CO2H have a greater nucleophilicity than CH3OH?

An orbital can contain up to 2 electrons, and the electrons and orbitals are described by wave functions, are the 3 wave functions the same? for example, two electrons that are in the psi 2 0 0 orbital do they have the same wave function as that orbital?

What does "air is saturated with water" or "air is saturated with something" and how can I relate this to vapor pressure?

What is the physical meaning of enthalpy?

The volume that 6 moles occupy 134.4l and 12 moles occupy 268.8 moles, since 1 mol = 22.4l, for this the gas needs to expand

What is the difference between the work done by a gas and the workflow that the same gas performs to exist? why does the variation in the internal energy of the equation dH = dQ + nRT not consider the work done by the gas to exist (working flow) but consumes only the heat transfer? In the question below, the author considers dU = Q. He performs the calculation of the work done by the gas, separated from the internal energy (with an equation dH = dU + nRT ). If you know that in systems with constant volume dU = Q, but the gas volume varies (From 6 moles to 12 moles of gas), this does not imply an internal energy variation equal to dU = w + q (As stated in the 1st law of thermodynamics) and why consider dU = w + q and not dU = Q **? In practice or in what he did, the logic he did not use was. Question: “A constant volume calorimeter showed that the fuel loss in burning 1 mole of glucose is equal to -2559kJ in 298K, that is dU = -2555KJ”

Thank you. Thank you.

A variation of enthalpy at constant pressure is numerically equal to the variation of heat in a chemical reaction, but what is enthalpy in itself? what is its difference for internal energy, is it calculated as H = U + PV, or does that mean physically?

Guys, why in physics does the work done by the system have the POSITIVE sign, but in chemistry does the work done by the system have the NEGATIVE sign?

Why does the equation τ = -n.R.T.ln⁡ (v_2 / v_1) only work for reversible processes?