(1) A piston-cylinder device initially contains 0

(1) Example Mass of Air in a Room

Determine the mass of the air in a room whose dimensions are 4m5m6m at 100kPa and 25.

Solution The mass of air in a room is to be determined.

Analysis Air at specified conditions can be treated as an ideal gas. The gas constant of air is Rg=287kJ/(kgK) and the absolute temperature is T=25+273=298K.

The volume of the room is

V=4m5m6m=120m3

The mass of air in the room is determined from the ideal gas relation to be

(2) Example Boundary Work during a Constant-Volume

A rigid tank contains air at 500kPa and 150. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65 and 400kPa, respectively. Show this process on the p-V diagram. Determine the boundary work done during this process.

Solution

The boundary work can be determined to be

This is expected since a rigid tank has a constant volume and dV= 0 in the above equation. Therefore, there is no boundary work done during this process. That is, the boundary work done during a constant-volume process is always zero. This is also evident from the p-V diagram of the process (the area under the process curve is zero).

(3) Example Cooling of a Hot Fluid in a Tank

A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800kJ. During the cooling process, the fluid loses 500kJ of heat, and the paddle wheel does 100kJ of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel.

Solution Take the contents of the tank as the system.

This is a closed system since no mass crosses the boundary during the process. We observe that the volume of a rigid tank is constant and thus there is no boundary work and V2=V1 . Also, heat is lost from the system and shaft work is done in the system.

Assumption The tank is stationary and thus the kinetic and potential energy changes are zero.

Therefore, and internal energy is the only form of the systems energy that may change during this process.

100kJ-500kJ = U2 - 800kJ , U2 = 400kJ

Therefore, the final internal energy of the system is 400kJ.

(4)

(5)

p, kPa

500

400

1

2

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