LECTURE
9
FIRST
LAW OF THERMODYNAMICS
The first law of thermodynamics is the
thermodynamic expression of the
conservation of energy.
This law most simply stated by saying that
“energy can not be created or destroyed” or that “the energy of the universe is
constant”.
This law can be stated for a system (control
mass) undergoing a cycle or for a change of state of a system.
Stated for a system undergoing a cycle, the
cyclic integral of the work is proportional to the cyclic integral of the heat.
Mathematically stated, for a control mass undergoing a cyclic
process such as in Joule’s experiment and for consistent set of units
∫dQfrom system= ∫dWon
system
or ∫dQfrom system- ∫dWon
system = 0
The important thing to remember is that the
first law states that the energy is conserved always.
Sign
convention The work done by a
system on the surroundings is treated as a positive quantity.
Similarly, energy transfer as heat to the system
from the surroundings is assigned a positive sign. With the sign convention one
can write,
∫dQ = ∫dW
Consequences
of the first law:
Suppose a system is taken from state 1 to state 2 by the path 1-a-2 and is
restored to the initial state by the path 2-b-1, then the system has undergone
a cyclic process 1-a-2-b-1. If the system is restored to the initial state by
path 2-c-1, then the system has undergone the cyclic change 1-a-2-c-1. Let us
apply the first law of thermodynamics to the cyclic processes 1-a-2-b-1 and
1-a-2-c-1 to obtain
∫1-a-2dQ+ ∫2-b-1dQ
- ∫1-a-2dW - ∫2-b-1dW =0
∫1-a-2dQ+ ∫2-c-1dQ
- ∫1-a-2dW - ∫2-c-1dW=0
Subtracting, we get
∫2b1dQ- ∫2c1dQ
–( ∫2b1dW - ∫2c1dW) =0
We know that the work is a path function and
hence the term in the bracket is non-zero. Hence we find
∫2b1dQ = ∫2c1dQ
That is heat is also a path function.
Energy
is a property of the system: By
rearranging we can have
∫2b1
(dQ - dW) = ∫2c1 (dQ - dW)
It shows that the integral is the same for the
paths 2-b-1 and 2-c-1, connecting the states 2 and 1. That is, the quantity
∫ (dQ - dW) does not depend on the path followed by a system,
but depends only on the initial and the final states of the system. That is ∫ (dQ - dW) is an exact differential of a property. This
property is called energy (E). It is given by
dE = dQ-dW
E = KE + PE +U
where U is the internal energy. Therefore,
dE = d(KE) + d(PE) + dU = dQ-dW
Quit often in many situations the KE or PE
changes are negligible.
dU = dQ – dW
An isolated system does not exchange energy with
the surroundings in the form of work as well as heat. Hence dQ = 0 and dW = 0.
Then the first law of thermodynamics reduces to dE = 0 or E2 = E1 that is energy of an
isolated system remains constant.
Perpetual
Motion Machine of the first kind: An imaginary device which delivers work continuously without absorbing
energy from the surroundings is called a Perpetual Motion machine of the first
kind. Since the device has to deliver work continuously, it has to operate on a
cycle. If such a device does not absorb energy from its surroundings ∫dQ =0. From the first law, it can
be observed that ∫dW =0, if
∫ dQ = 0. Therefore such a device is impossible from
first law of thermodynamics.
First
law analysis of non-flow processes: The first law of thermodynamics can be applied to a
system to evaluate the changes in its energy when it undergoes a change of
state while interacting with its surroundings. The processes that are usually
encountered in thermodynamic analysis of systems can be identified as any one
or a combination of the following elementary processes:
Constant volume (isochoric) process
Constant pressure (isobaric) process
Constant temperature (isothermal) process.
Adiabatic process.
Constant volume process: Suppose a gas enclosed
in a rigid vessel is interacting with the surroundings and absorbs energy Q as
heat. Since the vessel is rigid, the work done W due to expansion or
compression is zero. Applying the first law, we get
dU = dQ or Q = U2 –U1
That is, heat interaction is equal to the change
in internal energy of the gas. If the system contains a mass m equal of an
ideal gas, then
Q = ΔU = mCv (T2 –T1)
The path followed by the gas is shown on a P-V
diagram. Now consider the fluid contained in a rigid vessel as shown. The
vessel is rigid and insulated. Shaft work is done on the system by a paddle
wheel as shown in Fig. a. In Fig. b electric work is done on the system. Since
the vessel is rigid, the PdV work is zero. Moreover, the vessel is insulated
and hence dQ = 0. Application of the first law of thermodynamics gives
dU = dQ – dW = dQ – (dWpdv + dWs)
or dU = -dWs or – Ws =
ΔU = U2 –U1
Where dWpdv is the compression
/expansion work and dWs is the shaft work. That is increase in
internal energy of a system at a constant volume, which is enclosed by an
adiabatic wall, is equal to the shaft work done on the system.
Constant
pressure process: Several
industrial processes are carried out at constant pressure. A few examples of
constant pressure processes are: (a) reversible heating/cooling of a gas (b)
phase change (c) paddle wheel work (d) electrical work. For a constant pressure
process, the work done W is given by
W = ∫PdV = P (V2-V1)
Application of the first law of thermodynamics
gives
dU = dQ – dW = dQ – PdV
= dQ – d(PV)
or dQ = dU + d(PV) =
d(U + PV) = dH
or Q = ΔH
That is in a constant pressure process, the heat
interaction is equal to the increase in the enthalpy of the system. Now
consider the constant pressure processes in which the system is enclosed by an
adiabatic boundary. Application of the first law gives:
dU = dQ – dW = dQ –
(PdV + dWs)
Here, the net work done (dW) consists of two
parts – the PdV work associated with the motion of the boundary and (-dWs),
the shaft work (or electrical work) done by the surroundings. Since the system
is enclosed by an adiabatic boundary, dQ = 0 the equation can be written as
-dWs = dU + d(PV) = dH
That is, the increase in the enthalpy of the system is
equal to the shaft work done on the system.
Constant temperature process: Suppose a gas enclosed in the piston cylinder
assembly is allowed to expand from P1 to P2 while the
temperature is held constant. Then application of the first law gives:
dU = dQ – dW = dQ –PdV
It is not possible to calculate work and heat
interactions unless the relationships between the thermodynamic properties of
the gas are known. Suppose the gas under consideration is an ideal gas (which
follows the relation Pv = RT and u = u(T) only) then for an isothermal process,
dU = 0
dQ = PdV = RTdv/v or Q =W = RTln(v2/v1)
Reversible adiabatic
(Isentropic process):