First
Law of Thermodynamics:
The first law of thermodynamics, known as the law of
conservation of energy, states that the total energy of an isolated system
remains constant though it may change from one form to another.
When a system moves from state 1 to state 2, its internal energy
changes from U1 to U2. Then change in internal energy
ΔU = U2 – U1.
This internal energy change is brought about by the either
absorption or evolution of heat and/or by work being done by/on the system.
Because the total energy of the system must remain
constant, we can write the mathematical statement of the First Law as:
ΔU = q + w
–––––––––––––––––(7.7)
Where q - the amount of heat supplied to the system; w -
work done on the system
Other statements of first law of thermodynamics
1. Whenever an energy of a particular type disappears, an
equivalent amount of another type must be produced.
2. The total energy of a system and surrounding remains
constant (or conserved)
3. "Energy can neither be created nor destroyed, but
may be converted from one form to another".
4. "The change in the internal energy of a closed
system is equal to the energy that passes through its boundary as heat or
work".
5. "Heat and work are two ways of changing a system's
internal energy"
The mathematical statement of the first law of thermodynamics
is
∆U = q + w
--------- 7.7
Case 1 : For a cyclic process involving isothermal expansion of an
ideal gas,
∆U = 0.
eqn (7.7) ⇒∴ q = -w
In other words, during a cyclic process, the amount of
heat absorbed by the system is equal to work done by the system.
Case 2 : For an isochoric process (no change in volume) there is no work
of expansion. i.e. ΔV = 0
ΔU = q + w
= q - PΔV
ΔV =0
ΔU = qv
In other words, during an isochoric process, the amount of
heat supplied to the system is converted to its internal energy.
Case 3 : For an adiabatic process there is no change in heat. i.e. q= 0.
Hence
q = 0
eqn (7.7) ⇒ ΔU = w
In other words, in an adiabatic process, the decrease in
internal energy is exactly equal to the work done by the system on its
surroundings.
Case 4 : For an isobaric process. There is no change in the pressure. P
remains constant. Hence
∆U = q + w
∆U = q - P ∆V
In other words, in an isobaric process a part of heat
absorbed by the system is used for PV expansion work and the remaining is added
to the internal energy of the system.
Problem: 7.1
A gas contained in a cylinder fitted with a frictionless
piston expands against a constant external pressure of 1 atm from a volume of 5
litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy
from its surroundings. Determine the change in internal energy of system.
Solution:
Given data
q = 400 J
V1=5L
V2 = 10L
∆u = q-w (heat is given to the system (+q); work is done by the system(-w)
∆u q - PdV
= 400 J - 1 atm (10-5)L
=400 J - 5 atm L
[∴ 1 L atm = 101.33 J]
=400 J - 5 × 101.33 J
=400 J - 506.65 J
=- 106.65 J
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