TITLE.PM5

FIRST LAW OF THERMODYNAMICS 131

dharm

M-therm/th4-2.pm5

(i) Applying the first law energy equation,

Q = ∆ U + W

• 45 = ∆ U + (– 564)
  ∴ ∆U = 519 kJ. (Ans.)
  This shows that the internal energy is increased.
  (ii) Change in enthalpy,
  ∆ H = ∆ U + ∆ (pV)
  = 519 × 10^3 + (p 2 V 2 – p 1 V 1 )
  p 1 =^5
  V 1

• 1.5 =
  5
  0.15


• 1.5 = 34.83 bar
  = 34.83 × 10^5 N/m^2

p 2 =^5
V 2

• 1.5 = 005.^5 + 1.5
  = 101.5 bar = 101.5 × 10^5 N/m^2
  ∴∆ H = 519 × 10^3 + (101.5 × 10^5 × 0.05 – 34.83 × 10^5 × 0.15)
  = 519 × 10^3 + 10^3 (507.5 – 522.45)
  = 10^3 (519 + 507.5 – 522.45) = 504 kJ
  ∴ Change in enthalpy = 504 kJ. (Ans.)
  +Example 4.17. The following equation gives the internal energy of a certain substance
  u = 3.64 pv + 90
  where u is kJ/kg, p is in kPa and v is in m^3 /kg.
  A system composed of 3.5 kg of this substance expands from an initial pressure of 500 kPa
  and a volume of 0.25 m^3 to a final pressure 100 kPa in a process in which pressure and volume are
  related by pv1.25 = constant.
  (i)If the expansion is quasi-static, find Q, ∆U and W for the process.
  (ii)In another process, the same system expands according to the same pressure-volume
  relationship as in part (i), and from the same initial state to the same final state as in part (i), but
  the heat transfer in this case is 32 kJ. Find the work transfer for this process.
  (iii)Explain the difference in work transfer in parts (i) and (ii).
  Solution. Internal energy equation : u = 3.64 pv + 90
  Initial volume, V 1 = 0.25 m^3
  Initial pressure, p 1 = 500 kPa
  Final pressure, p 2 = 100 kPa
  Process : pv1.25 = constant.
  (i) Now, u = 3.64 pv + 90
  ∆ u = u 2 – u 1
  = 3.64 (p 2 v 2 – p 1 v 1 ) ...per kg
  ∴∆ U = 3.64 (p 2 V 2 – p 1 V 1 ) ...for 3.5 kg
  Now, p 1 V 1 1.25 = p 2 V 2 1.25

V 2 = V 1 p

p

1

2

F 1/ 25

HG

I

KJ

1.

= 0.25^500

100

1/ 25

F

HG

I

KJ

1.

= 0.906 m^3