- #1
Nylex said:The entropy change of the ideal gas isn't zero. You know that the change in internal energy is 0 for the gas (because U is a function of T only, so no change in T means no change in U) and from the first law therefore, you can get what the heat input to the gas is, in terms of the work done on/by the gas (and you see you have a numerical value, so you don't need to actually calculate anything) and hence calculate the change in entropy.
OlderDan said:Why is the entropy change of the gas not zero? There is no change of P, V or T. However you formulate dS (for constant n) it involves differentials of these three variables, so the "entropy content" of any state of an ideal gas can be determined. In this problem, isn't the gas just a conduit for the entropy flow into the surroundings with no entropy change of its own?
Of course the entropy leaving the gas is not the same as the entropy entering the surroundings because the temperatures are different. So "flow" should not be taken to mean the entropy is conserved.
Clausius2 said:I do think there is a change of entropy of gas. This problem is super-simplified, and maybe the professor looks for a simple solution like this:
[tex] \Delta S_{gas}=\frac{Q}{T_{gas}}=\frac{200}{273+40}[/tex]
[tex] \Delta S_{surroundings}=\frac{-200}{273+25}[/tex]
The work done over the gas is converted into heat which outflows the reservoir assuming an steady state ([tex]T_{gas}=constant[/tex]).
I do see your reasonement claiming that P,V, and T remains constant and so [tex]\Delta S_{gas}=0[/tex]. Well, this is not correct. Although the solution of this problem is a bit rudimentary (according with the rudimentary data provided), a more serious problem has to model an increasing of gas entropy. Just at the bottom of the logic of this problem is the effect caused both by the paddle wheel and the heat transfer. Both mechanisms involve entropy generation. When you are stirring the gas sure there are pressure and temperature gradients inside the gas, also assuming a semi-periodic steady state. In addition to that, the proper difference of temperatures between the gas and surroundings will generate entropy.
A complete formulation for this problem must be done via Fluid. Mech equations, in particular the entropy equation of a gas:
[tex] \frac{\partial}{\partial t}(\rho s)+\nabla\cdot(\rho\overline{v}s)=-\frac{\nabla\cdot \overline{q}}{T}+\frac{\phi}{T}[/tex]
where:
[tex]\overline{v}[/tex] velocity field caused by the stirring movement
[tex]\overline{q}[/tex] heat flux generated by temperature gradients
[tex]\phi[/tex] Rayleigh internal viscous dissipation function.
This three mechanisms will enhance a gas entropy generation. The last numbers I have made (and surely what the professor waits) are only a rudimentary model of this one.
tony_engin said:Um..as you have mentioned, there are two production of entropy geneartaoin..one from stirring the the gas and the other is the heat transfer.
The former one probably would produce entropy and the latter one seems to pass entropy to the surrounding. So will the two effects just cancel out so that the entropy change equals zero? This also seem to be justified by the two constant state properties (temperature and specific volume of the gas)
SpaceTiger said:I'm going to have to go with OlderDan and tony_engin on this one. Entropy is a state function, so if the gas is left in exactly the same state (T, V) as before the stirring, it ought to have the same entropy. You seem to know what you're doing, tony_engin, I'm not sure why you think you need our help.
Clausius2 said:I have given my opinion. If you super simplify the problem then you will be right. Anyway, I am (I am going to be in few time) an engineer, so calculating something which is far away from reality does not make sense to me. If this problem were real, surely it will be a change of entropy due to the equation I posted (reference: J. Spurk "Fluid Mechanics", Ed. Springer Verlag). If you want to stay with simplified stuffs, then go on.
SpaceTiger said:So I guess you think there's no value in learning Newtonian mechanics by neglecting frictional losses or air drag? Honestly, just suck it up and admit you were wrong. Don't pretend like you're above the problem and scoff at students trying to learn the fundamentals of thermodynamics.
Clausius2 said:What's wrong with you, boy?.
SpaceTiger said:You know I'm older than you, right? .
Clausius2 said:Anyway I don't know why you have felt so upset with me. I don't understand.
Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it is important because it helps us understand the direction and extent of energy transfer and transformations in a system.
Entropy can be calculated using the formula S = Q/T, where S is the change in entropy, Q is the heat transferred, and T is the temperature in Kelvin. This equation is based on the Second Law of Thermodynamics, which states that entropy always increases in a closed system.
The Ideal Gas Equation is PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. This equation is used to describe the behavior of ideal gases. Entropy is related to this equation through the Third Law of Thermodynamics, which states that the entropy of a pure, perfect crystal at absolute zero is zero.
According to the Second Law of Thermodynamics, entropy always increases in a closed system. However, in some cases, the entropy of a particular system may decrease while the overall entropy of the universe increases. This is possible through energy transfer and transformations in the system and its surroundings.
The change in entropy of a system can be used to predict the spontaneity of a process. If the change in entropy is positive, the process is likely to be spontaneous as it increases the disorder in the system. On the other hand, if the change in entropy is negative, the process may not be spontaneous as it decreases the disorder in the system.