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Can we use the volume to fix the thermodynamic state of a system?

If not, why do we use P-V curves to know about thermodynamic states and processes?

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    $\begingroup$ We can use specific volume (volume per mole or volume per unit mass) in conjunction with one other intensive property to fix the thermodynamic equilibrium state of a pure single-phase substance. $\endgroup$ – Chet Miller Aug 21 '18 at 12:56
  • $\begingroup$ Can we treat volume as intensive property for closed system(control mass)? $\endgroup$ – Kiran Dornala Aug 21 '18 at 13:30
  • $\begingroup$ In textbooks those i am referring to, displacement work for closed system is plotted using pressure and volume not specific volume.I am unable to understand why they took volume(extensive property) instead of specific volume(intensive property) to fix the state of the thermodynamic system.Is it okay to use volume for closed system to fix state? $\endgroup$ – Kiran Dornala Aug 21 '18 at 13:37
  • $\begingroup$ Sure, if you know the mass (either directly or from knowing the temperature and applying the ideal gas law). If the process is reversible, all the intermediate states will be thermodynamic equilibrium states. If the process is irreversible, only the initial and final states will be thermodynamic equilibrium states (assuming that the system is allowed to equilibrate at the final state). But, to get the work (which is not generally a function of state), you integrate ##P_{ext}dV## irrespective of whether the process is reversible. $\endgroup$ – Chet Miller Aug 21 '18 at 15:15
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As Chester Miller pointed out, volume alone does not fix the thermodynamic state of a system. The reason we use PV curves for closed systems is for determining total work. Insofar as using specific volume $v$ or total volume $V$ is concerned, since mass in a closed system is constant (not a variable) you normally use total volume $V$ to find total work done. For open systems where mass and/or mass flow rates may be variables, you would use specific volume in computing work per unit mass. For example, for an open system with steady mass flow rate we can obtain the rate of work (power) by multiplying the work per unit mass (kJ/kg) times the mass flow rate (kg/s).

Hope this helps.

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