A piston–cylinder device contains 0.87 kg of refrigerant-134a at –10°C. The piston that is free to move has a mass of 12 kg and a diameter of 25 cm. The local atmospheric pressure is 88 kPa. Now, heat is transferred to refrigerant-134a until the temperature is 15°C. Use data from the tables. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant-134a.

a) 90383 Pa; b) 0.0199 m^3; c) 4457 J

Explanation:

In this process pressure is applied from the atmosphere and the piston. As the piston can freely move, the pressure is constant throughout the process.  

Pressure, applied from the piston, can be calculated using the gravity force divided by the area of the piston. As both pressures have the same direction, total applied pressure is a sum of both- atmospheric and piston-based- pressures. This value can be calculated as:  

 p=patm+mg/(πd^2/4)=90398 Pa

b)

Using Ideal Gaseous Law, we can calculate initial and final volumes of the refrigerant. As the law is based on the amount of the substance n, we need to use molar mas s of the refrigerant equal to 102 g/mol. In addition, the temperatures should be converted to Kelvins, so the initial temperature is 263 K and the final temperature is 288 K.  

For the initial and final values, the following calculations can be done:  

V=mRT/(pM), then V1=0.206 m^3 and V2=0.2259 m^3

Then, the change in the volume is 0.0199 m^3

c)

Change in enthalpy can be calculated as the work plus the change in the internal energy. In the isobaric process, the work is equal to the pressure times change in volume. The change in the internal energy is calculated using the change in the temperature. Finally, the following calculations can be done:  

 H=W+U=p(V2-V1)+3/2*m/M*R(T2-T1)=4487J

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