TITLE.PM5

FIRST LAW OF THERMODYNAMICS 139

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/M-therm/Th4-3.pm5

Also, bb+^1 = γ ...(Given)
∴ d(loge p) + γd (loge v) = 0
Integrating, we get pvγ = constant.
Example 4.25. A 15 cm diameter vertical cylinder, closed by a piston contains a combus-
tible mixture at a temperature of 30°C. The piston is free to move and its weight is such that the
mixture pressure is 3 bar. Upper surface of the piston is exposed to the atmosphere. The mixture
is ignited. As the reaction proceeds, the piston moves slowly upwards and heat transfer to the
surroundings takes place. When the reaction is complete and the contents have been reduced to
the initial temperature of 30°C, it is found that the piston has moved upwards a distance of
8.5 cm and the magnitude of heat transfer is 4 kJ. Evaluate :
(i)The work ;
(ii)Decrease in internal energy of the system.
Solution. Diameter of vertical cylinder,d = 15 cm (or 0.15 m)
Temperature of combustible mixture = 30°C (or 303 K)
Pressure of the mixture = 3 bar = 3 × 10^5 N/m^2
Upward displacement of the system = 8.5 cm (or 0.085 m)
Magnitude of heat transfer, Q = – 4 kJ ...(i)
(i) Work done by the system, W = ∫ pdv
= 3 × 10^5 ∫ dv [Q p = constant = 3 × 10^5 N/m^2 ]

= 3 × 10^5

π

4

L ×× 0 2 0 085

NM

O

QP

() ..15 N-m

= 450.62 N-m or J = 0.4506 kJ

∴ W = 0.4506 kJ.

(ii) By first law of thermodynamics,

Q = ∆U + W

• 4 = ∆U + 0.4506
  ∴∆U = – 4.4506 kJ
  ∴ Decrease in internal energy = 4.4506 kJ. (Ans.)
  Example 4.26. A house wife, on a warm summer day, decides to beat the heat by closing
  the windows and doors in the kitchen and opening the refrigerator door. At first she feels cool
  and refreshed, but after a while the effect begins to wear off.
  Evaluate the situation as it relates to First Law of Thermodynamics, considering the
  room including the refrigerator as the system.
  Solution. Initially, the temperature of air in the room falls when it communicates with the
  cool refrigerator with its door open. This makes the house wife feel cool.
  Considering the room and its contents as the system, and assuming the walls, windows and
  doors non-conducting, we find, Q = 0.
  To operate the refrigerator, electricity is supplied from outside and hence external work W
  is done on the system.
  Applying the first law to the system,
  Q = ∆ U + W
  0 = ∆ U + (– W)
  ∴∆ U = W