The enthalpy is the actual internal energy of the system plus the part of the energy of the environment that is an honorary part of the internal energy because conceptually you had to do work PV on the environment (assuming constant P) to create a space for the system to exist. If you want to change volume of the system, you of course have to do work on the system itself, but you also have to do work on the environment to make it accommodate the different-sized system, and it’s the total of these that the difference in the enthalpy measures.
What's the relation between Enthalpy, Internal Energy, Entropy, and Total Energy?
The question should be more specific. H = U+PV For reversible isotermal process, Q = TΔS Energy balace, ΔE = total energy ΔE = ΔU + ΔEK + ΔEP + ΔE_electric + .... = Q - W Assume system simple incompesible system, with constant pressure process, ΔU = Q - W, W = PΔV ΔU = TΔS - PΔV ΔH = TΔS --> Equilibrium process.
Enthalpy vs. Internal Energy?
The internal energy refers to energy of the molecules internal to the substance, therefore it's another form of energy. In contrast, enthalpy is "a definition for conveniece". The term "U + PV" is used so much in thermodynamcis that it is given its own name: Enthalpy (H). It is defined as follows: H = U + PV You can conclude from this formula that the enthalpy and the internal energy of a substance (system) cannot have the same value. (unless P = 0 or V = 0) In practice, we seldom use the absolute values for U and H. Rather, the changes in these two properties are important for us, say delta(H) and delta(U) delta(H) = delta(U) + delta(PV) a change in the enthalpy of a system is equal to the change in internal energy of the same system added to the change in P*V.
Enthalpy and internal energy! Help, please ;)?
ΔH = ΔE + Δng RT ΔH - ΔE = ΔngRT where Δng = no.of moles of gaseous products - no.of moles of gaseous reactants now more the Δng whether positive or negative more the difference between ΔH and ΔE A. Δng = 1 - 0 = 1 B.Δng = 1 - 0 = 1 C.Δng = (1 + 1) - ( 1 + 1) = 2 - 2 = 0 D.Δng = 4 - ( 2 + 7) = 4 - 9 = -5 E. (4 + 6) - ( 4 + 5) = 10 - 9 = 1 as Δng is highest for D ....so D is the answer feel free to ask any question
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is 5084.3 kJ.?
_____Work and heat dU = Q + W 5084.3 kJ = 5074.3 kJ + W SOLVE Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem.
Help calculating the internal energy of a gas?
The only thing you need to know is the temperatures to calculate the change in internal energy. Internal energy is supposed to be independent of pressure in the case of an ideal gas. In the case of a calorically perfect gas, the change in internal energy is proportional to the change in temperature: Change in specific internal energy: ∆u = c_v*(T2 - T1) Even though the process is not constant volume, the constant volume specific heat applies, and is in fact defined off the concept of internal energy. I will show how to get c_v from the adiabatic index, universal gas constant, and molar mass: The definition of adiabatic index: gamma = c_p/c_v The relation between enthalpy & internal energy: h = u + P*v For an ideal gas: h = u + R/M * T If specific heats are constant: c_p = c_v + R/M Solve this with gamma = c_p/c_v. Our end results are: c_p = R/M * gamma/(gamma - 1) c_v = R/M * 1/(gamma - 1) Thus: ∆u = R/M * 1/(gamma-1) *(T2 - T1) multiply by mass to be applicable for the bulk situation: ∆U = m*R/M * 1/(gamma-1) *(T2 - T1) Data: R:=8.314 J/mol-K; M:=28.96 g/mol; T2:=450 K; T1:=279.2 K; gamma:=1.4; m:=2 g; (you've stated wrong in the above for m) Result: ∆U = 245.17 Joules --------------------------------------... Actually, if you didn't make a typo in the first paragraph and the mistake is in the assessment of states, the process is actually at constant volume. --------------------------------------... If you do want to know how to set up the integral, you can assume that it takes a polytropic form. This means: P1*V1^n = P2*V2^n, and it progresses though all intermediate states under this condition. Solve for n, and then use the formula for the work done. The integral has already been done for you. W = (P2*V2 - P1*V1)/(1 - n) In the special case that n = gamma, the process is adiabatic. If it isn't, use the first law of thermodynamics to calculate the heat transferred. Q = ∆U + W
How come (change in internal energy = change in KE +change inPE+ change in enthalpy ) while .....?
i guess you must understand the meaning of every term in the equation. It is difficult if you just rely on the mathematics of it without understanding the physical meaning of it. in thermodynamics, our concern is to get the relationship between energy and work. enthalpy = internal energy + PV if the substance is rigid such as solid, it will not change pressure or volume PV=0 dEnthalphy =dInternal energy But for gases and fluids, some of the energy is used to expand or contract itself, therefore we need to specify this energy as PV. internal energy is the sum of kinetic energy (if it is hot) and potential energy ( if chemically reactive) U= KE+ PE. but it is difficult to get the absolute internal energy, so what we do is to get the change in internal energy delta U, by measuring energy that we put in as heat , Q or the work it has done to surroundings such that dU=Q+W Therefore: H= U + PV H= (KE + PE) + PV dH = dU + d(PV) dH= Q+W + d(PV)
Meaning of enthalpy and entropy?
Lancenigo di Villorba (TV), Italy QUESTION n.1 "Can someone explain to me the meaning of enthalpy and entropy?" ENTROPY APPEARED IN SECOND HALF OF XIX CENTURY BY CLAUSIUS PAPER. Entropy is a Thermodynamics Property ; ITS IMMEDIATE MEANING RELATES TO DISORDER STATUS OF Thermodynamic Systems. ENTHALPY APPEARED IN ENDINGS OF XIX CENTURY BY GIBBS PAPER. Enthalpy is a Thermodynamics Property ; ITS IMMEDIATE MEANING RELATES TO WHOLE ENERGY TAKEN INTO A Thermodynamic Systems. QUESTION n.2 "What is the relationship between these two and the reaction rate of a system?" GIBBS HIMSELF GET HIS "DEFINED FREE ENERGY EQUATION" Delta(G) = Delta(H) - T * Delta(S) HAVING MEANING FOR ISOTHERMAL PROCESS. ANOTHER SCIENTIST, J. E. VAN'T HOFF ASSIGNED "CHEMICAL EQUILIBRIUM" FOR Thermodynamic Systems ONCE HE STARTED BY Delta(G) = 0 Delta(H) = T * Delta(S) AT THESE CONDITIONs, VAN'T HOFF LOOKED A PARTICULAR RELATIONSHIP AMONG Enthalpy AND Entropy. Entropy AND Enthalpy BELONG TO THERMODYNAMICS WHILST Rata Reaction BELONGS TO KINETICS. KINETICS TREATS UPON REACTION SPEED (e.g. evoluting conditions) WHILST THERMODYNAMICS TREATS UPON CHEMICAL EQUILIBRIA (e.g. Equilibria are conditions undefinitely kept on basis of time). I hope this helps you.