In thermodynamics, the interaction whose external system could be viewed as the raising of mass through a distance against gravitational force is defined as work done by a system on the surroundings during a process.

**Thermodynamics Formulas** are listed below.

Some thermodynamic functions cannot be calculated straightforwardly. One needs to be able to present these quantities regarding others that can be experimentally determined.

**Solved Examples**

**Problem 1:** Calculate how much heat is either added or removed from the system when 100kJ of work is done on a closed system during a process and the total energy of the system increases by 55.0kJ.?

**Answer**:

By the energy conservation principle, a gross energy transfer to system effects in an equivalent increase of internal energy stockpiled in the system. This may be articulated as

Q = DeltaE + W

This equation is typical statement of first law. It says that in any alteration of state the heat supplied to a system is equal to the work finished by the system plus the upsurge of internal energy in the system.

Bearing in mind the work is done on a system as positive

Q + (+100.0) = +55.0

Q = +55.0 – 100.0

Q = -45.0kJ

From the result owing to the negative sign 45.0kJ of energy in heat form is removed from the system all through the procedure.

**Problem 2**: For a pound of Freon 12 refrigerant calculate the work done during compression. The refrigerant is at 20 psia and 30oF and is compressed to 140 psia and 150oF during a compression stroke.

**Answer**:

From the table of thermodynamic properties for refrigerent 12 the unique and concluding specific volumes are 2.0884 and 0.33350 cu ft/lb respectively. For the procedure

frac14020 = left(frac2.08840.33350right)2

7 = (6.262)n

n = 1.0607

The work done in compression =

W = fracP2V2–P1V11−n

W = frac1441−1.067 times (140 times 0.3335-20 times 2.0884)

W = -11,670ft-1b