TITLE.PM5

FIRST LAW OF THERMODYNAMICS 215

dharm

/M-therm/Th4-7.pm5

(ii)The piston will be in virtual equilibrium and hence zero work is effected by
the piston. (Ans.)
(iii) Work done by oxygen = work done on methane and expansion of oxygen is effected in the
system
∴ Woxygen = + 24.7 kJ
and Q = (U 2 – U 1 ) + W

Amount of oxygen present ==

××

××

pV

RT

11

1

4 105 035

0 1000 333

.5 0.

.24

= 0.197 kg

and T 2 =

pV

pV

22

11

× T 1 =

11.47 07 333

4

××

×

0.

.5 0.035

= 1697.5 K. (Ans.)

(As the piston is free, the final pressure of oxygen and methane will be same).

∴ Q = (U 2 – U 1 ) + W

= mcv (T 2 – T 1 ) + W

= 0.197 × 0.64 (1697.5 – 333) + 24.7 = 196.7 kJ. (Ans.)

Highlights

1. Internal energy is the heat energy stored in a gas. The internal energy of a perfect gas is a function of
   temperature only.


2. First law of thermodynamics states :
   — Heat and work are mutually convertible but since energy can neither be created nor destroyed, the
   total energy associated with an energy conversion remains constant.
   Or
   — No machine can produce energy without corresponding expenditure of energy, i.e., it is impossible to
   construct a perpetual motion machine of first kind.
   First law can be expressed as follows :
   Q = ∆E + W
   Q = ∆U + W ... if electric, magnetic, chemical energies are absent and changes in
   potential and kinetic energies are neglected.


3. There can be no machine which would continuously supply mechanical work without some form of energy
   disappearing simultaneously. Such a fictitious machine is called a perpetual motion machine of the first
   kind, or in brief, PMM1. A PMM1 is thus impossible.


4. The energy of an isolated system is always constant.


5. In case of
   (i)Reversible constant volume process (v = constant)
   ∆u = cv(T 2 – T 1 ) ; W = 0 ; Q = cv (T 2 – T 1 )
   (ii)Reversible constant pressure process (p = constant)
   ∆u = cv(T 2 – T 1 ) ; W = p(v 2 – v 1 ) ; Q = cp (T 2 – T 1 )
   (iii)Reversible temperature or isothermal process (pv = constant)
   ∆u = 0, W = p 1 V 1 loge r, Q = W
   where r = expansion or compression ratio.
   (iv)Reversible adiabatic process (pvγ = constant)

± ∆u = + W = RT T()^12 −γ− 1 ; Q = 0 ; TT^2 vv pp

1

1

2

1

2

1

1

=F

HG

I

KJ

=F

HG

I

KJ

γ− γ−γ