Question

A mass of 0.07 kg of air at a temperature of 30°C and pressure 100 kPa is compressed to a pressure of 600 kPa, according to the law:

(i.e. Polytropic process, n=1.3). Determine:

i.The final volume

ii.The final temperature

iii.The work transfer

iv.The change in internal energy

v.The heat transfer

Answer 1

Ans -

Mass of air, m = 0.07 kg

Initial temperaure, T1 = 30 oC = 303 K

Initial pressure, P1 = 100 kPa

Final pressure, P2 = 600 kPa

(i) Considering air as an ideal gas, we have

P1V1 = mRT1

100 * V1 = 0.07 * 0.287 * 303 [ For air, R = 0.287 kJ/kg K ]

V1 = 0.061 m3  

According to the law,  

P1V1n = P2V2n

100 * 0.0611.3 = 600 * V21.3

  V2 = 0.015 m3 (ans)

(ii) We know, that

T2 = 458.16 K = 185.16 oC     (ans)

(iii) Work transfer for a polytropic process is given by

  

(ans)

negative sign indicates that the work is done on the system

(iv) For air, specific heat at constant volume is, cv = 0.718 kJ/kg K

Therefore, the change is internal energy is,

  

     (ans)

positive sign indicates that the internal energy increases

(v) According to the first law of thermodynamics, we have

  

where is the heat transfer

Therefore,

(ans)

negative sign indicates that the heat is transferred into the system