Calc based physics thermodynamics

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The integral of W = -∫Pdv depends on the specific process path, such as isobaric, isochoric, isothermal, or isentropic. For a constant pressure process, the boundary work simplifies to W_b = P(V_2 - V_1). To perform the integration, pressure must be expressed as a function of volume, which can be derived from the ideal gas law. Understanding the type of thermodynamic process is crucial for accurately calculating work done. Properly defining the relationship between pressure and volume is essential for solving these integrals in thermodynamics.
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hi. i was wondering on how to do this integral. W= - integral Pdv. what is the integral of dv ?
 
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lukasz08 said:
hi. i was wondering on how to do this integral. W= - integral Pdv. what is the integral of dv ?

It depends on the process path. If you assume it is a constant pressure process, then the boundary work is just W_b = P(V_2 - V_1).

CS
 
You integrate the pressure with respect to volume. Depending on the process (isobaric, isochoric, isothermal, isentropic) this function (p = f(v)) will be different. These functions can be derived using the ideal gas law.
 

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