First law of thermodynamics for

closed system
 Learning Objectives
1. Examine the moving boundary work or P dV work commonly
encountered in reciprocating devices such as automotive engines
and compressors.
2. Identify the first law of thermodynamics as simply a statement of
the conservation of energy principle for closed (fixed mass)
systems.
3. Develop the general energy balance applied to closed systems.
4. Define the specific heat at constant volume and the specific heat at
constant pressure.
5. Relate the specific heats to the calculation of the changes in internal
energy and enthalpy of ideal gases.
6. Describe incompressible substances and determine the changes in
their internal energy and enthalpy.
7. Solve energy balance problems for closed (fixed mass) systems that
involve heat and work interactions for general pure substances,
ideal gases, and incompressible substances.
 Moving boundary work

• Form of mechanical work which is

commonly observed

• Associated with piston cylinder

device where the inner face of the
piston (part of system boundary)
moves back and forth, thus, is
called moving boundary work

• Is also called 𝑃𝑑𝑉 work

• Commonly encountered in
automobile engines, compressors
e.t.c.
 Moving boundary work

• For real engines(compressors):-

– The process is very fast consequently, the process
path can not be specified

– Work, being a path function, cannot be determined

analytically without a knowledge of the path

– Thus, the boundary work can not be predicted

exactly from thermodynamics alone.
 Moving boundary work

Quasie-quilibrium process,

• A process during which the system

remains nearly in equilibrium at all
times, the boundary work is:-

• The total B.W. during the entire process

is:-
 Moving boundary work

+ Ve during expansion process , sine ∆V is +Ve

• The B.W.
- Ve during compression process, since ∆V is - Ve

• If , which is the equation of the process path on a

P-V diagram, is known, can be obtained from the integral
 Moving boundary work

• Since work is a path function,

different paths followed by the
system b/n two states associates
with different work

• A cyclic process will have a net

work out put
 Moving boundary work

EXAMPLE 4–1
• A rigid tank contains air at 500 kPa and 150°C. As a result of
heat transfer to the surroundings, the temperature and pressure
inside the tank drop to 65°C and 400 kPa, respectively.
Determine the boundary work done during this process.

EXAMPLE 4–2
• A frictionless piston–cylinder device contains 10 lbm of steam
at 60 psia and 320F. Heat is now transferred to the steam until
the temperature reaches 400F. If the piston is not attached to a
shaft and its mass is constant, determine the work done by the
steam during this process.
 Moving boundary work

EXAMPLE 4–3
• A piston–cylinder device initially contains 0.4 m3 of air at 100
kPa and 80°C. The air is now compressed to 0.1 m3 in such a
way that the temperature inside the cylinder remains constant.
Determine the work done during this process.
 Moving boundary work

Polytropic Process

• A process by which P and V of a

gas during compression and
expansion processes are related
by:-

• For Polytropic process will

be:-
 Moving boundary work

EXAMPLE 4–4
• A piston–cylinder device contains 0.05 m3 of a gas initially at
200 kPa. At this state, a linear spring that has a spring constant
of 150 kN/m is touching the piston but exerting no force on it.
Now heat is transferred to the gas, causing the piston to rise
and to compress the spring until the volume inside the cylinder
doubles. If the cross-sectional area of the piston is 0.25 m2,
determine
– (a) the final pressure inside the cylinder,
– (b) the total work done by the gas, and
– (c) the fraction of this work done against the spring to
compress it.
 Energy balance for closed system
• Energy balance for any system undergoing any kind of
process:-

• In rate from

• Per unit mass basis

• In differential form
 Energy balance for closed system

• For a closed system undergoing a cycle:-

• Thus the energy balance becomes

• The general energy balance equation considering the sign

convention ( the first law of thermodynamics) will be:-
 Energy balance for closed system
The first law cannot be proven mathematically, but no process
in nature is known to have violated the first law, and this
should be taken as sufficient proof

EXAMPLE 4–5
• A piston–cylinder device contains 25 g of saturated water
vapor that is maintained at a constant pressure of 300 kPa. A
resistance heater within the cylinder is turned on and passes a
current of 0.2 A for 5 min from a 120-V source. At the same
time, a heat loss of 3.7 kJ occurs. (a) Show that for a closed
system the boundary work Wb and the change in internal
energy U in the first-law relation can be combined into one
term, H, for a constant pressure process. (b) Determine the
final temperature of the steam.
 Energy balance for closed system

EXAMPLE 4–6
• A rigid tank is divided into two equal parts by a partition.
Initially, one side of the tank contains 5 kg of water at 200 kPa
and 25°C, and the other side is evacuated. The partition is then
removed, and the water expands into the entire tank. The water
is allowed to exchange heat with its surroundings until the
temperature in the tank returns to the initial value of 25°C.
Determine
– (a) the volume of the tank,
– (b) the final pressure, and
– (c) the heat transfer for this process.
 Specific heat
• It is the energy required to raise the
temperature of unit mass a substance by
one degree

• This energy depends on how the

processes is executed

• If the process is executed as the volume

is maintained constant , it will be
specific heat at constant volume

• If the process is executed as the pressure

is maintained constant , it will be
specific heat at constant pressure
 Specific heat
• > always. But why?
Specific heats in terms of other thermodynamic properties:-
– consider a fixed mass in a stationary closed system undergoing
a constant-volume process

– From differential form of energy conservation principle

– From definition of

or
 Specific heat
• Similarly, an expression for can be obtained by considering a
constant-pressure expansion or compression process. Drive it!

• Since and are expressed in terms of other properties, they

must also be properties themselves.

Some points about and

• Are property relations and thus are independent of the type of

processes

• is a measure of the variation of internal energy of a substance

with temperature, and is a measure of the variation of enthalpy
of a substance with temperature
 Specific heat

• Both internal energy and enthalpy of a substance can be

changed by the transfer of energy in any form (not by heat
only) thus, specific energy is probably more appropriate than
the term specific heat
Internal energy, Enthalpy, and Specific heats of ideal gases

• For an ideal gas the internal energy is

a function of the temperature only

• Through his experiment, Joule

showed Internal energy is a function
of temperature only and not a
function of pressure or specific
volume

• Enthalpy of an ideal gas is also

function of temperature only
Internal energy, Enthalpy, and Specific heats of ideal gases

• Thus for an ideal gas

• At low pressures, all real gases approach ideal-gas behavior,

and therefore their specific heats depend on temperature only

• The specific heats of real gases at low pressures are called

ideal-gas specific heats ( and )
Internal energy, Enthalpy, and Specific heats of ideal gases
• ways to determine the internal energy and enthalpy changes of
ideal gases

– By using the tabulated u and h data. This is the easiest and

most accurate way when tables are readily available.

– By using the cv or cp relations as a function of temperature

and performing the integrations. This is very inconvenient
for hand calculations but quite desirable for computerized
calculations. The results obtained are very accurate.

– By using average specific heats. This is very simple and

certainly very convenient when property tables are not
available. The results obtained are reasonably accurate if
the temperature interval is not very large.
Internal energy, Enthalpy, and Specific heats of ideal gases

• Specific Heat Relations of Ideal Gases

But = and

• Replacing dh by cp dT and du by cv dT and dividing the

resulting expression by dT gives:
Internal energy, Enthalpy, and Specific heats of ideal gases

EXAMPLE 4–7
• Air at 300 K and 200 kPa is heated at constant pressure to 600
K. Determine the change in internal energy of air per unit
mass, using (a) data from the air table (Table A–17), (b) the
functional form of the specific heat (Table A–2c),and (c) the
average specific heat value (Table A–2b).

EXAMPLE 4–8
• An insulated rigid tank initially contains 1.5 lbm of helium at
80°F and50 psia. A paddle wheel with a power rating of 0.02
hp is operated within the tank for 30 min. Determine (a) the
final temperature and (b) the final pressure of the helium gas.
 Internal energy, Enthalpy, and Specific heats of ideal gases
EXAMPLE 4–9
• A piston–cylinder device initially contains 0.5 m3 of nitrogen gas at
400 kPa and 27°C. An electric heater within the device is turned on
and is allowed to pass a current of 2 A for 5 min from a 120-V
source. Nitrogen expands at constant pressure, and a heat loss of
2800 J occurs during the process. Determine the final temperature
of nitrogen.

EXAMPLE 4–10
• A piston–cylinder device initially contains air at 150 kPa and 27°C.
At this state, the piston is resting on a pair of stops, as shown in Fig.
4–32, and the enclosed volume is 400 L. The mass of the piston is
such that a 350-kPa pressure is required to move it. The air is now
heated until its volume has doubled. Determine (a) the final
temperature, (b) the work done by the air, and (c) the total heat
transferred to the air.
Internal energy, Enthalpy,& Specific heats of solids & liquids

• Liquids and solids can be approximated as

incompressible substances without
sacrificing much in accuracy

• The constant-volume and constant-pressure

specific heats are identical for
incompressible substances

• Like those of ideal gases, the specific heats

of incompressible substances depend on
temperature only
Internal energy, Enthalpy,& Specific heats of solids & liquids

• Thus the internal energy will be obtained as:

• For small temperature intervals, a c value at the average

temperature can be used and treated as a constant
Internal energy, Enthalpy,& Specific heats of solids & liquids
• For enthalpy

• For solids, the term is insignificant and thus

• For liquids, two special cases are commonly encountered:

I. Constant-pressure processes, as in heaters :

II. Constant-temperature processes, as in pumps :

Internal energy, Enthalpy,& Specific heats of solids & liquids

EXAMPLE 4–12
• A 50-kg iron block at 80°C is dropped into an insulated tank that
contains 0.5 m3 of liquid water at 25°C. Determine the temperature
when thermal equilibrium is reached.

EXAMPLE 4–13
• If you ever slapped someone or got slapped yourself, you probably
remember the burning sensation. Imagine you had the unfortunate
occasion of being slapped by an angry person, which caused the
temperature of the affected area of your face to rise by 1.8°C
(ouch!). Assuming the slapping hand has a mass of 1.2 kg and
about 0.150 kg of the tissue on the face and the hand is affected by
the incident, estimate the velocity of the hand just before impact.
Take the specific heat of the tissue to be 3.8 kJ/kg · °C.