Problem solving tips

## Thermodynamics

- Want to score better marks in exams? Have a look at some problem solving tips.
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Tip-1: Conversion of work into heat
In many processes we need to calculate the energy liberation. Some of these processes are - dropping a stone from some height, passing a meteor through atmosphere, rotation of stirrer in a liquid etc. In all these processes some amount of work is involved. If we calculate the work done then we can easily calculate the equivalent amount of heat liberated by using the equation , where is the work done, H is the equivalent heat and J is the Joule's equivalent of heat. The value of is 1 in SI system but its value is erg/cal.
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Tip-2: Recognizing intensive and extensive variables
i) If X and Y are two intensive variables then , , , are also intensive.
ii) If X and Y are two extensive variables then is also extensive but , are intensive variables.
iii) If X is intensive and Y is extensive then , and will be extensive.
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Tip-3: Writing equation of state of a thermodynamic system
If X, Y and Z are thermodynamic coordinates specifying an equilibrium state of the system then the equation of state is written as . Following are the equation of state for some common physical systems-
Here, P =Pressure, V = Volume, T = Temperature, L =Length, F = Force, T = Tension, M = Magnetization, H = Magnetic field intensity, Z = Electrochemical equivalent, E = Emf, A = Area, S =Surface tension
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Tip-4: Calculating work done
In order to calculate work done in a process, we have to first identify the process (isothermal, adiabatic, isobaric or isochoric) and then use the expression . Following table will be useful to calculate the work done in various process.
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Tip-5: Entropy changes for an ideal gas
If we want to calculate the entropy change in a reversible process using an ideal gas then we can utilize any of the following form depending on the nature of the process.
i) when temperature and volume changes:
ii) when temperature and pressure changes:
iii) when pressure and volume changes:
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Tip-6: Identifying problems on heat engine and refrigerator
Both Heat Engine and Refrigerator operate within two heat baths (one source and one sink). But their operation is completely opposite. If the system is taking heat from the hot bath (or source) and releasing some part of heat to the cold bath (or sink) then the system performs some work. This system acts as a heat engine.
On the other hand when the system takes heat from the sink and release some part of it to the source then work needs to be done on the system. This system acts as a refrigerator. Therefore it is important to identify whether the system is a heat engine or a refrigerator.
In case of heat engine we calculate the efficiency and in case of refrigerator we calculate coefficient of performance. Following trick can help you distinguish a heat engine and a refrigerator.