If heat ‘q’ is supplied to a system, it may be used up partly to increase the internal energy of the system and partly to produce some mechanical work. If DE is increase in internal energy of the system and ‘w’ is the work done by the system, then we will have,

**q =**

**D**

**E + w**

This equation is the mathematical formulation of first law of thermodynamics. It may be written in a number of other forms as follows;

DE = q – w.

If the changes are very small, it may be written as;

dE = dq – dw.

If the work done is the work of expansion only and the volume increases by a small amount dV against the external pressure P, then we know that dw = PdV

Therefore,

**E = q – PdV.** From the above equation it is clear that at constant volume i.e. dV = 0,

*“The change in internal energy is equal to heat absorbed or evolved by the system.”*
1st law of thermodynamics is the law of conservation of energy. If you ask about the usual form of this law in thermodynamics, the elementary answer is following. Suppose you have a gas with internal energy EE. Now, you bring some heat into the gas. We usually work with infinitesimally small quantities, so I denote this heat by dQdQ, where dd means “infinitesimally small”. Thanks to this additional heat, the internal energy EE increases by dEdE. If nothing else happens, we have dE=dQdE=dQ, i.e. all the heat we injected into the system is used to increas the internal energy. However, usually the increase of internal energy is accompanied by some work done by the gas. For example, the gas will expand a little bit and it will displace some object, like the cover of the vessel inside which the gas is. In that case, the heat dQdQ will be be used to increase the internal energy and to perform the work dAdA. So, the total balance is

ReplyDeletedE=dQ−dAdE=dQ−dA

which means: the total increase of internal energy is equal to the heat brought to the system minus the work done by the gas. This discussion should be supplemented by the usual remark that EE is a state variable and hence dEdE is an exact differential, while heat and work are not state variables, and their differentials are not exact.

In our particular case we can easily find the expression for dAdA. Recall that the elementary volume dV=dSdx,dV=dSdx,where S is the area of the base and dxdxis infinitesimal displacement. On the other hand, pressure exerted by force dFdF on the area element dSdS is defined by P=dFdS.P=dFdS.Work is defined as force times displacement, i.e. dA=Fdx.dA=Fdx. Putting all this together, we find

dA=dFdx=PdSdx=PdV.dA=dFdx=PdSdx=PdV.

Hence, the 1st law of thermodynamics is often written in the form

dE=dQ−PdV.dE=dQ−PdV.

For reversible processes, the heat can be related to the entropy SS (not to be confused with the area, it’s the same symbol for different quantity), but this is not so easy (some integrability conditions must be investigated and analyzed), so thath

dE=TdS−PdV,dE=TdS−PdV,

where TT is the temperature.

So, if you were asking how this particular formula is derived, this is the brief answer.

Thanks for yor explanation.

DeleteWhy do you double de symbol respersentation EE, QQ, AA and in the equations you use the single representations E, Q, A?

Thanks

ReplyDelete