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## Thermodynamic Quantities Defined

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Thermodynamic Quantities Defined

Internal Energy = U = the sum of all the energy held by the molecules:

* the PE stored in their chemical bonds, attractions and repulsions

* the KE of their motion; translation, rotation, vibration

Heat = Q = is what flows between bodies when one loses internal energy, and another gains it.

U1

@ T1

U2

@ T2

Q

Work can be done on a gas.

U = +

Internal energy increases

U = Won

But the side effect of the man doing work is that the gas might get hotter and some of the heat leak out of the walls.

U = Won - Q lost

We can prevent this from happening by insulating the piston and chamber so that no heat can go in or out (adiabatic means Q = 0, no heat flows)

Work can be done by a gas.

U = neg

Internal energy is used up (lost, decreases) as piston pushes up.

U = - Wby

But the side effect is that the gas might cool and some of the heat be sucked into the walls.

U = - Wby+ Q into

We can prevent this from happening by insulating the piston and chamber so that no heat can go in or out (adiabatic means Q = 0, no heat flows)

The First Law of Thermodynamics

The total change in the internal energy of a system is the sum of the work done on the system and the heat transferred to the system.

U = Won + Q into

If Q = 0 (adiabatic) , then ΔU is positive if positive work is done on the system.

If W = 0, (isochoric: box cant expand or contract), then ΔU is positive if heat flows into the system

Expansions and contractions of gases are shown on PV diagrams

The area under a PV graphs has special significance.

Pressure of gas on walls

Volume of gas increases as pressure on walls decreases

Volume of gas decreases as pressure on walls increases

Typical of an expansion

Typical of an compression of gas

The area under a PV graphs has special significance.

x

Wgas = F x = PA x

But A x = volume compressed

Wgas = P V = area under a PV graph

Wgas = P V = area under a PV graph

Try calculating the work done by the gas in this isobaric expansion

P = 1.01 x 105 Pa , Vi = .7 m3, Vf = 1.3 m3

W = P V = (1.01 x 105 N/m2)( .6 m3)= 60600 J

What if the arrow were switched and it was an isobaric compression?

W = P V = (1.01 x 105 N/m2)(- .6 m3)= - 60600 J

Net work = 0

+W

-W

Isobaric expansion

How much work is done?

W= PV=Po (3Vo) = 3PoVo

Name the process B to C

Isochoric loss of pressure

How much work is done?

W= PV=Po 0 = 0

Name the process C to A

Contraction

How much work is done?

W= PV=can’t be done because P is changing

W = estimate of area under curve = 4.5 boxes

4.5 boxes (1 box = ½ PoVo) = -2.25 PoV0

Net Work done in cycle = 3PoVo +0 + -2.25 PoVo= + .75 PoVo

Net Work done in cycle = 3PoVo +0 + -2.25 PoVo= + .75 PoVo

Do you see a shortcut?

Get the area of the enclosed triangle

W= ½ bh = ½ 3Vo (½ Po) = ¾ PoVo

So for any closed cycle, the net work done is the area enclosed .

For an open cycle (where you don’t return to the P, V, T you started at) the work done is the sum of the areas under the curve

V

On each GRAPH below, draw lines and curves to indicate 3 processes: isobaric, isochoric and isothermal.V

P

T

T

P

P

V

V

V

Closed Cycles: WHEN A SYSTEM RETURNS TO ITS SAME P and VSince PV = nRT, that means it returns to the same temperature as well. U = 0, and the first law reduces to ___________.

This means W = - Q and all work done is lost as ________. The work done is also the area of the enclosed cycle.

Tell how the first law changes for process that are

Adiabatic

Isothermal

Isobaric

Isochoric

U = W + Q

U = W + Q

U = W + Q

U = W + Q

Another powerful tool: if you know PV of a gas, you can automatically calculate its temperature.

PV = nRT

# of moles gas constant

Find temperatures at points 1 and 2:

Assume the data below applies to a 20 mole sample of ideal gas

P1V1 = nRT1

So T1 =P1V1/nR= 5e5 (1)/20(8.3) = 7600°K

P2V2 = nRT2

So T2 =P2V2/nR= 2e5 (.5)/20(8.3) = 60°K

Interpretation: the gas _____________ and ____________

Isothermal = at constant temperature

PV = nRT = a constant

What’s the shape of the curve xy = 1 ?

This shape is called an isotherm . You must actually know T at all points or calculate it to be sure; can’t tell by shape alone.

Occurs in an ice bath (thermal resevoir) that can exchange heat with walls if temperature starts to change

When a gas expands adiabatically, the work done in the expansion comes at the expense of the internal energy of the gas causing the temperature of the gas to drop. The figure below shows P-V diagrams for these two processes.

U = Won + Q into

Which process resulted in a higher temperature?

Thus the adiabat lies below the isotherm.

In the end, the internal energy of a gas depends only on its temperature, assuming PVT changes only.

Chemical or phase changes could change PE of molecules, but we don’t deal with that in this course.

U = KEint + PE int = (3/2) nkT

Example: One mole of monatomic ideal gas is enclosed under a frictionless piston. A series of processes occur, and eventually the state of the gas returns to its initial state with a P-V diagram as shown below. Answer the following in terms of P0, V0, and R.

- Find the temperature at each vertex.
- Find the change in internal energy for each process.
- Find the work by the gas done for each process.

Example: One mole of ideal gas is at pressure P0 and volume V0. The gas then undergoes three processes.

- The gas expands isothermally to 2V0 while heat Q flows into the gas.
- The gas is compressed at constant pressure back to the original volume.
- The pressure is increased while holding the volume constant until the gas returns to its initial state.
- A. Draw a P-V diagram that depicts this cycle. Label relevant points on the axes. In terms of T0, the initial temperature, label each vertex with the temperature of the gas at that point.
- For the remaining sections, answer in terms of T0, Q, and R.
- B. Find the change in internal energy for each leg of the cycle. C. Find the work done by the gas on each leg of the cycle. D. Find the heat that flows into the gas on legs 2 and 3. E. Find the efficiency of this cycle.

Example: One mole of ideal gas is at pressure P0 and volume V0. The gas then undergoes three processes.

- The gas expands isothermally to 2V0 while heat Q flows into the gas.
- The gas is compressed at constant pressure back to the original volume.
- The pressure is increased while holding the volume constant until the gas returns to its initial state.
- A. Draw a P-V diagram that depicts this cycle. Label relevant points on the axes. In terms of T0, the initial temperature, label each vertex with the temperature of the gas at that point.
- For the remaining sections, answer in terms of T0, Q, and R.
- B. Find the change in internal energy for each leg of the cycle. C. Find the work done by the gas on each leg of the cycle. D. Find the heat that flows into the gas on legs 2 and 3. E. Find the efficiency of this cycle.

For solution go to

http://apcentral.collegeboard.com/members/article/1,3046,151-165-0-44428,00.html

Example: Calculate the internal energy of the air in a typical room with volume 40 m3. Treat the air as if it were a monatomic ideal gas at 1 atm = 1.01 105 Pa.

You can use the gas law PV=nRT to express the internal energy in terms of pressure and volume.

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