Period 7 Thermodynamics #3 by Thomas E Clark

Tamera

2017-02-08 22:33:01 UTC

Pg 358
1) The first law of thermodynamics deals with energy conservation within a system.
2) In order for mechanical energy to be conserved, the system needs to be closed so no energy is lost to heat.
Pg 359
1) ΔU=3000 J -2000 J=1000 J
2) ΔU=2500 J -1700 J= 800 J
Pg 360
1)The only way a gas can do work is if the gas expands (and pushes against its surroundings/ container).
2) For gases: W= PAd.
Pg 361
1) A process is known to be isobaric when the pressure stays constant.
2) In an isobaric process, the volume of the system changes while the pressure stays constant.
Pg 362
1) W= 1000 Pa (300 m^3- 50 m^3)= 250,000 J
2) 6000 J= 300 Pa (x- 100)
x= 120 m^3
Pg 363- 364
1) When the pressure is not constant in a system and the volume is constant, it is known as a isochoric.
2) The area under the graph tells you the work that has been done.
Pg 364
1) Since the system does not have a change in volume, this means that there was no work done.
W= 3000 Pa (0)= 0 J
2) Since the system does not have a change in volume, this means that there was no work done.
Pg 365-366
1) When there is a isothermal system, the temperature remains constant while other measurements change.
2) PV= nRT; In an isothermal process, the internal energy does not change.
Pg 367
1) W= (1 mol)(8.31 J/K)(303.15 K)(ln(1.5))= 1020 J
2) W= (0.6 mol)(8.31 J/K)(298.15 K)(ln(3))= 1630 J

Komal Mittal

2017-02-08 22:34:31 UTC

pg 358
1. Internal energy = heat - work
2. Q is positive when the system absorbs heat, W is positive when the system does work on the surroundings.

pg 359
1. U = 3,000 + 2,000 = 5,000 J
2. U = 2,5000 + 1,700 = 42,000 J

pg 360
1. W = P * delta V
2. Graph of pressure v work shows changes in relation to each other, and the area under the curve represents work.

pg 361
1. Isobaric processes occur when pressure stays constant.
2. Graphically, pressure is represented by a horizontal line.

pg 362
1. W = (1000)(300-50) = 250000 J
2. 6000 = (300)(x-100)
Final V = 120 cubic m

pg 363-364
1. Isochoric processes occur when the volume is constant.
2. No work is done during an isochoric process.

pg 364
1. W = 0 J (no change in volume, no work)
2. W = 0 J (no change in volume, no work)

pg 365-366
1. Isothermal processes are when there is no change in temperature.
2. W = nRTln(Vf/Vi)

pg 367
1. W = (1)(8.31)(273.15+30)ln (3/2) = 1,020 J
2. W = (.60)(8.31)(273.15+25)ln(3/1) = 1630 J

Jordan and Sitara

2017-02-08 22:34:37 UTC

pg 358
1. First law of thermodynamics says energy (heat, work, and internal energy) is conserved.
2. When a system does work on its surroundings and gives off no waste heat, its internal energy, U, changes by W.
pg 359
1. U=3000-2000=1000J
2. U=2500-1700= 800J
pg 360
1. Gas only performs works only if the gas expands (W=Fs)
2. W=PAs
3. W=P*change in volume
pg 361
1. When pressure stays constant, its called isobaric
2. isobaric system can change in volume but still have a constant pressure (adding heat while expanding volume)
pg 362
1. 1,000Pa*(300m^3-50m^3)= 250000 J
2. 6000J=300Pa(x-100m^3)
x=120m^3
pg 363-364
1. volume is constant: isochoric process
2. Since volume is constant in isochoric process, no work is done
pg 364
1. 0J because there is no change in volume
1. 0J because there is no change in volume
pg 365-366
1. isothermal system: temperature remains constant as other quantities change.
2. W=nRTln(Vf/Vi)
pg 367
1. W=(1 mol)(8.31 J/K)(273.15K+30K)ln(3/2)= 1020 J
2. W =(.60 mol)(8.31 J/K)(273.15K+25K)ln(3/1)= 1630 J

Sakshi, Diana, Catlan

2017-02-08 22:35:41 UTC

Pg 358 2 Key Points
1. The first law of thermodynamics states that energy is conserved.
2. Uf-Ui=Change U=Q-W
Uf=final internal energy
Qi=initial internal energy in a system
Q= heat is absorbed or relased by the system
-W= system does work on the surroundings
Pg 359 Practice Questions
1. 3000J - 2000J = 1000J
2. 2500J - 1700J = 800J
Pg 360 2 Key Points
1. A gas performs work only if the gas expands and it shown by W=Fs
s=distance.
2. The area times distances equals change in volume.
W=P*Chanve V
Pg 361 2 Key Points
1. Isobaric is when a process has the pressure stay constant.
2. W=P Change V
Pg 363 2 Key Points
1. Volume is constant, then its a isochoric process.
2. If the volume is constant, then force times distance equals zero.
Pg 362 Practice Questions
1. 1000Pa(300m^3-50m^3) = 250,000J
2. 6000J = 300Pa(x-100m^3)
20 m^3 = x - 100m^3
x = 120 m^3
Pg 363 key pts
1. Isochoric process is a constant volume.
2. Because the volume is constant, no work is being done.
Pg 364 Practice Questions
1. 0 J (No change in volume)
2. 0 J (No change in volume)
Pg 365 key pts
1. Isothermal system have constant temperature but other quantities can change.
2.PV=nRT
Pg 366 key pts
1.Work is done by
2. W=nRTln(final volume/initial volume)
Pg 367 Practice Questions
1. W = (1 mol)(8.31 J/K)(273.15 K + 30K)ln(3/2) = 1,020J
2. W = (0.60mol)(8.31 J/K)(273.15 K + 25 K)ln(3/1) = 1,630 J

Caitlin and Alex

2017-02-08 22:39:22 UTC

Key Points pg 358:
1. A form of energy involved is the internal energy that's in the motion of atoms and molecules.
2. The first law of thermodynamics is important since it ties all the quantities together.
Practice pg 359:
1. U = 3000 J - 2000 J = 1000 J
2. U = 2500 J - 1700 J = 800 J
Key Points pg 360:
1. A gas begins working if only the gas expands.
2. W = Fs
Key Points pg 361:
1, An isobaric process is when the pressure stays constant.
2. In the image, the volume of the gas is changing and then the piston is what keeps the pressure constant.
Practice pg 362:
1. W = P*V
(1,000 Pa)(300 m^3 - 50 m^3) = 250000 J
2. Vf = W/P + Vi
=( 6,000 J/ 300 Pa) + 100 m^3= 120 m^3
Key Points 363-364:
1. Having the pressure in the system isn't constant, one may see that a simple closed container in which the volume can't be changed.
2. In the figure, the volume is constant meaning that Fs = 0.
Practice pg 364:
1. It is 0 J because there isn't change in the volume meaning the process is isochoric.
2. It's also 0 J because there isn't a change in volume and no work is being done.
Key Points pg 365-366:
1. The temperature is constant as the quantities change which means it's an isothermal system.
2. Having PV = nRT means the temperature stays constant.
Practice pg 367:
1. W = nRT ln(Vf/Vi)
= (1.0 mole)(8.31 J/K)(273.15 K + 30 K) ln(3.0/2.0 cubic m) = 1020 J
2. (.60 moles)(8.31)(273.15 K + 25 K) ln(3.0/1.0) = 1633.171 J

Bonny

2017-02-08 22:42:04 UTC

pg 358
1.Change in Internal Energy of a system= Heat absorbed/released by system - the work the system does on its surroundings
2. W>0 when system does work on the surroundings
W<0 when the surroundings do work on the system

pg 359
1.Q=3000J W=2000J
Change in motor's internal energy= 3000J-2000J=1000J
2. Q=2,500J W=1,700J
Change in motor's internal energy= 2,500J-1,700J=800 J

pg 360
1. W=Forcexdistance and since F=Pressurexarea, Work=PressurexAreaxDistance and AreaxDistance= Change in volume, so Work=PressurexChangeinVolume
2. There are four standard conditions under which work is performed in thermodynamics: constant pressure, constant volume, constant temperature, and constant heat.

pg 361
1. A process that is isobaric means that the pressure is constant.
2. There can be a situation where the volume is changing, but the pressure is not changing because a weighted piston keeps the pressure constant.

pg 362
1. Change in Volume= 300-50=250 cubic meters
Pressure= 1000Pascals
W=(250m^3)(1000Pascals)= 250,000J
2. P=300Pascals, W=6000J
Initial volume=100m^3
W=(P)(Vfinal-Vinitial)
6000J=(300Pascals)(Vfinal-100m^3)
Final volume=120m^3

pg 363-364
1. An isochoric process is when the volume in a system is constant.
2. Because volume is constant in an isochoric process, no work is done because ΔV=0 and W=PΔV.

pg 364
1. ΔV=0 and W=PΔV, so the gas did 0J of work regardless of the pressure.
2. ΔV=0 and W=PΔV, so the gas did 0J of work regardless of the pressure.

pg 365-366
1. An isothermal system means that the temperature remains constant in the system.
2. In an isothermal process, the work done is W=(nRT)(ln(Vf/Vi)) where R, the gas constant=8.31J/molK.
3. In an isothermal system, Q=W or in other words, the heat equals the work done by the system.

pg 367
1. n=1mol T=30+273.15K Vf=3.0m^3 Vi=2.0m^3
W=(1mol)(8.31J/molK)(30+273.15)(ln(3/2))= 1,021 J
2. n=0.6mol T=25+273.15K Vf=3.0m^3 Vi=1.0m^3
W=(0.6mol)(8.31J/molK)(298.15K)(ln3/1)= 1,633J and since W=Q, Q= 1,633J,

Rachel and Chisondi

2017-02-08 22:42:26 UTC

Page 358
-Internal energy resides in the motion of the atoms and molecules (vibrations and them sliding over each other)
-The first law of thermodynamics deals with the conservation of energy, tying all of the quantities (heat, work and internal energy) together
Page 359
1. Delta U = 3000J -2000J= 1000J
2. Delta U = 2500J-1700= 800J
Page 360
-A gas preforms work only as it is expanding
-W= Fs = PAs = PDeltaV
Page 361
-Isobaric is a process where pressure stays constant
-Isobaric CAN have a change in volume (without a change in pressure necessarily)
Page 362
1. W = 1000P x (300 cubic meters - 50 cubic meters) = 250000J
2. W = 6000J/300 Pa + 100 cubic meters = 120 cubic meters
Page 363-364
-An isochoric process volume is constant but pressure is not
-No work is done in an isochoric process
1. 0 J because no change in volume
2. 0 J because no change in volume
Page 365-366
-isothermal process the temperature stays constant while other factors change
-area under the curve represents the work done in an isothermal process
Page 367
1. W=nRTln(Vf/Vi)= (1.0)(8.31J/K)(273.15K + 30.0K)ln(3.0/2.0)=1,020 J

Ruth

2017-02-08 22:44:01 UTC

pg.358 - 2 key points
1) For chemical energy to be conserved you have to work with systems where no energy is lost to heat - there could be no friction.
2) These three quantities - heat, work, and internal energy - considered to make up all the energy
pg. 359
1)1000J
2) 800J
Pg. 360
1) A gas performs work only if the gas expands w=Fs
W= PAs, force equals pressure times area which means that you can write work as pressure times area times distance.
2) W= P delta V, the area times the distance (As) = the change in volume, delta V
pg 361
1)

Jose Ibarra

2017-02-08 22:45:24 UTC

Pg.358
- The quantity W(work) is positive when the system does work on its surroundings and negative when the surroundings do work on the system. (balloon inflating and deflating)
- The first law of thermodynamics states that these energies are compressed.
Pg.360
-If a gas is doing work while you keep its temperature constant, the amount of work performed and the intermediate and final states of the system will be different from when pressure was constant.
-When anything changes in the process the change is assumed to be quasi-static meaning the change gets slow enough that the pressure and temp. are the same throughout the system's volume.
Pg.361
-isobaric is when volume changes but pressure stays the same.
-W=P^V
Pg.363-364
-If volume is constant in an isochoric process there's no work.
-isochoric process has a constant volume.

Hiwi and Yohe

2017-02-08 22:46:36 UTC

Key points: pg. 358
- For mechanical energy to be conserved, you have to work with systems where no energy is lost to heat.
- The system's energy decreases if deltaU is negative.
Pg. 359
1) U= 3000J-2000J=1000J
2) U=2500-1700J= 800J
Key points: pg. 360
- A gas performs work only if the gas expands.
- The curve shows how pressure and volume change in relation each other.
Key points: pg. 361
-Isobaric means when you have a process where the pressure stays constant.
- An isobaric system may feature a change in volume, but the pressure remains constant.
Pg 262 Q
1) W=PdeltaV
w=(1,000Pa)(300m^3-50.m^3)= 250,000J
2) Vf=W/P+Vi
Vf= 6,000/ 300+100m^3
=120m^3
Key Points: Pg. 363-364
-When volume is constant you have an isochoric process.
-No work is done when volume is constant in an isochoric process.
Pg. 364
1)It is 0J because the volume does not change, so no work is done.
2) It is 0J because the volume does not change, so no work is done.
Key Points: Pg 365 and 366
- In an isothermal system, the temp. remains constant as other quantities change.
- The internal energy does not change because the temperature stays constant in an isothermal process.
Pg. 367
1) W=nRTIn(Vf/Vi)
(1.0)(8.31)(273.15K)+(30K) ln(3.0/2.0)
=1,020J
2) (.6Mole)(8.31J/K)(73.15K+25K)(3.0/1.0)
=1,630J