# thermo 01 first law and non flow process

Document Sample

```					Lecture 5: work, the first law of thermodynamics and non-flow processes Work done by an expanding system • the sketch shows a system trapped within a cylinder by a piston. The force on the piston F=pA where A is the piston area and p the pressure.

F

dx
System boundary
• •

the small amount of work done as the piston moves a distance dx is : dW = pAdx = p dV where V is the volume of the system. If the system expands from 1 to 2 then

W = ∫ pdV , w = ∫ pdv
1 1

2

2

•

this important result shows that the work done by a system as it expands is given by the area under the process path on a p-v or p-V diagram.

1

p

2

w = ∫ p dv
1

2

v

•

The first law of thermodynamics in short, the first law of thermodynamics states that ‘energy is conserved’. There are different ways to state this. For example, for a thermodynamic cycle, “If a system undergoes a cycle, then the net energy transferred into the system during the cycle as heat is equal to the net energy transferred out as work.”

•

consider the cycle

p
B

C 1

A 2

v

•

noting the convention that heat transfer into the system is positive and work out from the system is positive, the first law then states, ∫ dq = ∫ dw thus

•

∫
•

2 1A

dq + ∫ dq = ∫ dw + ∫ dw
2B 1A 2B

1

2

1

and

∫
•

2 1A

dq + ∫ dq = ∫ dw + ∫
2C 1A 1 1

1

2

1 2C

dw

if we subtract this last equation from the previous one, we then get

∫
•

2B

( dq − dw) = ∫ ( dq − dw)
2C

i.e. the change in (q –w) does not depend on the process path, only the end states, and so can be considered to be a change in a thermodynamic property. we can thus write

•

q − w = ∆e & Q − W = ∆E

•

•

∆e and ∆E can be considered to be changes in the ‘stored energy’ in the system. This energy can be considered to be composed of ‘internal energy’ u or U, as well as kinetic and potential energies. If we ignore the kinetic and potential energies for now, we can then write q − w = ∆u & Q − W = ∆U these last equations are sometimes called the ‘non-flow energy equations’ (NFEEs), and are one form of the first law of thermodynamics that we will use extensively.

Perpetual motion machines of the first kind • a machine that breaks the first law of thermodynamics is called a perpetual motion machine of the first kind. Joule’s Experiment • consider the following experiment, with two pressure vessels connected by a valve and in a bath of water Thermometer

Water Bath
•

phigh

plow

if the valve is opened, the thermometer shows no change provided that the substance is a perfect gas.

•

so, if we consider the contents of the pressure vessels as the system, Q=W=0, and thus ∆U = 0 from this, we can conclude Joule’s law: “The internal energy U depends only on temperature, and not pressure or volume, provided that the system is a perfect gas.”

•

•

Specific heats the specific heat of a substance is the amount of energy required to raise the temperature of 1 kg of the substance by 1°C. For most gases and vapours, which expand when heated, the specific heat depends on how the substance is heated. the specific heat at constant volume cv is found by keeping the volume of the system constant during heating. i.e. during an isochoric process, ⎛ ∂u ⎞ cv = ⎜ ⎟ ⎝ ∂T ⎠ v similarly, the specific heat at constant pressure cp is found by keeping the pressure of the system constant during heating. i.e. during an isobaric process. We will discuss this term later on.

•

•

Processes • let us consider a unit mass of some pure substance (the ideas can be applied to any system).
•

the following ideal processes involve changing the state of the system while holding one property constant. In each process there is firstly a general case, for which one would have to resort to charts or tables. Then follows the special case of a system consisting of an ideal gas.

The isochoric (constant volume) process
1

p

2
•

the work is

v
w = ∫ p dv = 0
1 2

•

thus, from the NFEE

q − w = ∆u ⇒ q = ∆u

• •

or, for a small step in the process

dq = du from the definition of cv above, cv ( dT )v = ( du ) v , thus
dq = cv dT

•

draw a three dimensional graph of u against v and T. Then cv at any point on the surface is the slope of that graph in the T direction:

u v

Constant v line

T
•

so far no particular substance has been assumed so this applies to any system. The change in internal energy could be found from tables. however, if the system is a perfect gas, then Joule’s law applies and u depends only on T and we no longer use a partial derivative above du cv = dT for constant specific heats, we can integrate this expression to get u = cv T + const

•

•

U = m cvT + const
•

usually we use T in K and set the constant to 0. This puts an arbitrary origin into our scale for internal energy. This is not normally a problem because the first law is interested in differences in internal energy, not absolute values. u = cvT ,

∆ u = cv ∆T
• note: 1. the conditions under which these relations hold: ideal gases with constant

specific heats only.
2. in this case, since u depends only on T, the above equations are true for any

process, not just an isochoric process.

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 243 posted: 7/19/2009 language: English pages: 4
How are you planning on using Docstoc?