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What is the role of partial derivative in thermodynamics
Partial derivative is an important mathematical tool to solve thermodynamic problems. Theoretically, the magnitude of all macroscopic physical effects can be expressed by the partial derivative of a state function to a state parameter. For example, in the throttling process, we use (DT / DP) h to express the degree of temperature change (increase or decrease) with the decrease of pressure, The partial derivative can be used for quantitative analysis
If you have any questions, please ask
5mol ideal gas, initial state 300K, 400kPa, constant pressure heating to 500K, calculate the volume work of the system to the environment?
…… I don't know how to ask for environmental pressure,
Why is the ambient temperature of constant pressure heating always 400k, and the system pressure here is the same as the ambient pressure
The ambient pressure of constant pressure heating is always 400kPa
You can take a constant pressure heating model in accordance with the initial state. For example, a light piston with an area of S is placed at the opening of a rectangular container, and the light piston moves △ X
Work w = PS △ x = P △ V
Assuming that N2 is an ideal gas, 2 DM3 N2 is used for isothermal expansion to P at 0 ° and 5p,
(1) Reversible expansion (2) expansion takes place at constant external pressure P
Calculate both Q w △ u △ H
The ideal gas equation is used to calculate the final state volume, isothermal, u = 0. U = q + W.W = rtlnv2 / v1
University physicochemical thermodynamics
Why is the enthalpy change greater than zero when a severe chemical reaction occurs in the isolation system, which makes the temperature and pressure of the system increase significantly? It is required that the enthalpy change should not be explained by the enthalpy change formula, but by reason
First floor: is this an endothermic reaction with rising temperature?
Second floor: please do not use formula to explain, but from the truth
It is easy to infer from the question that the volume of the isolation system will not decrease during the process, and the system does external work. However, the decrease of the volume does not change the conclusion
2. The process is obviously irreversible. It is easy to know that the work of the reversible process is large (the pressure of the reversible process is always greater than or equal to the external pressure of the irreversible process)
3. The △ (PV) = integral sign [D (PV)] = integral sign [PDV] + integral sign [VDP] of the system is always greater than the work done by the reversible process: the integral sign [PDV], so it must be greater than the external work done by the irreversible process (W)
4. The external work done by the isolation system (W) is equal to the reduction of internal energy (- Δ U). It is known from 3 that △ U + △ PV > 0
5. From the definition of enthalpy, △ H = △ U + △ PV, so the enthalpy change is greater than zero
Attachment: it's impossible for you to ask for an explanation without the formula of enthalpy change, because if you want to discuss the enthalpy change, if you don't know what enthalpy is (the definition of enthalpy), it will be impossible to discuss it. Your idea is very good. Indeed, we should try our best to understand the essence of the problem, not just from the formula to the formula, but it is sometimes troublesome to completely leave the formula and express the problem in pure natural language, It is not as concise as formula, and formula can be quantified, but natural language is difficult to quantify
In addition, what is the definition of isolation system you mentioned? Is it an isolated system or an adiabatic system or just a closed system (there is no material exchange with the outside world, but active power or heat exchange is allowed). If the first case is very simple, the work discussed above is zero, and the conclusion remains unchanged. If the third case is, it is assumed that most of the heat released by the reaction is dissipated to the outside world, The enthalpy of the system will decrease. So I'll deal with it in the second way
Thermodynamics of Physical Chemistry——
The volume of a container is 162.4 cubic meters, and there is air with pressure of 94430 PA and temperature of 288.65 K. when the container is heated to TX, the gas escaping from the container accounts for 114.3 cubic meters under pressure of 92834 PA and temperature of 289.15 K; then the value of TX is: () (a) 1038.15 K (b) 948.15 K (c) 849.15 K (d) 840.15 K
According to the formula of PV = NRT, we can get the initial n total = p1V1 / RT1, the gas overflowing from the container n escape = P2V2 / rt2, then the remaining gas quantity in the container n stay = N Total - N escape, then the gas temperature in the container TX = p1V1 / N leave R (because the pressure of the overflowing gas is the same as that in the container, otherwise it will not be balanced) into TX = p1V1
When to use the definition and the derivation formula to calculate the partial derivative at one point?
Generally, the definition is complicated and the derivation formula is used. Only the problems that cannot be solved by the derivation formula are considered by the definition
How to calculate the partial derivative?
Refer to college calculus 2, for example
z=2x^2+y^2
Dz/Dy=2y
Dz/Dx=4x
Those who seek x regard y as a natural number, while those who seek y regard x as a natural number
Second order partial derivative of implicit function,
I don't know how to get the right formula for the second partial derivative of X. some steps seem to be omitted in the middle. Can you help me write it out,
In the second step, we regard Z as a function of X. first, we find the derivative of Z, and then multiply it by the partial derivative of Z to X,
It is a derivative rule and chain rule of quotient
When to use the definition method to find the second order partial derivative
There are two methods to calculate partial derivative, one is definition method, the condition is generally discontinuous at a certain point, and the other is to calculate partial derivative limit (a variable remains unchanged, the number of another variable substituted into this point, such as X partial derivative at point (x0, Y0), the molecule is f (x, Y0) form), the condition of this method is that the function is continuous at this point
How to calculate the second order partial derivative with the first order partial derivative?
Let u = f (x, y) have a second order continuous partial derivative, and x = RCOs θ, y = rsin θ, find ᦉ 8706; 2U / ᦉ 8706; R2, I just can't understand how to find the second order partial derivative, how to use the first order partial derivative to find the second order partial derivative?
∂u/∂r = (∂u/∂x)(∂x/∂r) + (∂u/∂y)(∂y/∂r) = (∂u/∂x)*cosθ + (∂u/∂y)*sinθ, ∂²u/∂r...
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