# data[].attributes.body
Pg.358 2 Key Points
1. Energy can neither be created nor destroyed.
2.Heat is a transfer of thermal energy. Work is the transfer of mechanical energy.
Pg.359
1. 3000J - 2000J= 1000J
2. 2500J - 1700J= 800J
Pg.360 2 Key Points
1. Work = force * distance
or W=F * s
2. Work = change in volume * pressure
Pg.361 2 Key Points
1. Isobaric process is a thermodynamic process where pressure stays constant.
2. The pressure of the piston stays constant, but the volume of a gas can change.
Pg. 362
1. W=Pressure * change in volume
(1000 Pa)(300m^3 -50m^3)= 250000J
2. 6000J = (300 Pa) (final volume - 100) = 120m^3
Pg.363 2 Key Points
1. Isochoric process is a process where pressure is not constant but volume is constant.
2. An example of an isochoric process is if a spray can was tossed into a fire.
Pg.364
1. Since change in volume is 0, there is no work.
2. Since change in volume is 0, there is no work.
Pg.365-366 2 Key Points
1. In an isothermal system, temperature remains constant.
2. The relation between pressure and volume is
P= nRT/V
Pg.367
1. W= (1)(8.31J/K)(303.15K)ln(3/2)
= 1020 J
2. W= (.6)(8.31 J/K)(273+15)(ln 3) =1633 J

# data[].attributes.body
2) Graphing pressure versus volume graph compares them and the area under gives you work.
pg.361
1) An isobaric process is a thermodynamic process in which pressure stays constant meaning the change of P is equivalent to zero. The volume of the gas can change but the weighted piston maintains the continuity of the pressure.
2) Constant pressure, temperature change can change volume.
pg.362
1) W=(1000)*[(300M^s)-(50m^s)]=250000J or 250kj
2. 6kj-6000J=(300Pa)(Vf-100)=20
Vf-100=120m^3
pg.363-364
1) When volume is constant while there is a change in temperature then the pressure changes but there isn't any work performed.
2)  An isochoric process involves the pressure of a gas increasing, but not the volume.
pg.364
1)W= 3000*0=0
2) W= 6000*0=0
pg.365-366
1) In an isothermal system, the temperature remains constant as other quantities change.
2)P=nRT/V
pg.367
1.W=(1mol)*(8.31J/K)*(273K+30K)*[ln(3/2)]=1020J
2. W=(.6mol)*(8.31J/K)*(273K+25K)*[ln93/1)]=1630J

# data[].attributes.body
358
1) Energy can't be created or destroyed.
2) Work is the transfer of mechanical energy. Internal energy can change by doing work or a change in heat.

359
1) 3000J-2000J=1000J
2) 2500J-1700J=800J

360
1) W = PdeltaV

361
2) With constant pressure, temperature change can change volume.
2) isobaric = constant pressure

362
1) W = (1000 Pa)(300 m^3 - 50 m^3) = 250000 J
2) 6000 J = (300 Pa)(Vfin - 100)
20 = Vfin - 100; Vfin=120m^3

363-364
1) An isochoric process involves the pressure of a gas increasing, but not the volume.
2) Volume is constant in an isochoric process, no work is done.

364
1) No work, deltav=0
2) No work, deltav=0

365-366
1) Temperature is always constant in an isothermal process.
2) P=(nRT)/V

367
1) W=(1.0mol)(8.31J/K)(303.15K)ln(3/2)
W=1020J
2) W = (0.6mol)(8.31J/K)(273.15K + 25K)ln(3/1) =16

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358
1. First Law of Thermodynamics deals with conservation of energy. -internal energy persists in situations of motion of atoms and molecules.
2. Work is the transfer of mechanical energy-for work in a gas, it must be compressed. Mechanical energy-work with systems where no energy lost in heat.

359
1. Change of U- 3000J-2000J=1000J
2. 2500J-1700J=800J
360 and 361
1. The packet relates how Force=Pressure
W=P*(Change of V)
Work depends on the changes or consistency of volume as well as the pressure for a gas.
2. F=Pressure*Area
The graph builds the connection between the two, but the area underneath explains the amount of work in that current scenario.
3. An isobaric process is a thermodynamic process in which pressure stays constant meaning the change of P is equivalent to zero. The volume of the gas can change but the weighted piston maintains the continuity of the pressure.
4. Heat added to change the temp is represented by an equivalent value of work which was done by the gas.

362
1. W=(1000)*[(300M^s)-(50m^s)]=250000J or 250kj
2. 6kj-6000J=(300Pa)(Vf-100)=20
Vf-100=120m^3

363 and 364
1.  An isothermal process is measured as a changing system, but temperature remains constant. In other words, internal energy doesn't change while heat stays constant.
2. Volume a constant number, so work is zero.

364
1. V=0 therefore no work is done in this scenario. 3000*0=0
2. There is no change in volume.

365 and 366
1.W=nRT*ln(Vf/Vi) isothermal temp-constant_(P and V are constant)
2.P=nRT/V

367
1.W=(1mol)*(8.31J/K)*(273K+30K)*[ln(3/2)]=1020J
2. W=(.6mol)*(8.31J/K)*(273K+25K)*[ln93/1)]=1630J

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358
1.) Energy can neither be created nor destroyed.
2.) If you have two of heat, work, or internal energy, you can solve for the third and they all interact proportionally. Their sum is also finite.
359
1.)deltaU = 1000 J
2.)deltaU = 800J
360
1.) W = PdeltaV
2.)When you hold temperature constant, and change the pressure, the resultant change in volume of a gas is all that really matters in terms of it's work done.
361
1.)When you hold pressure constant, and change the temperature, the resultant change in volume of a gas is all that really matters in terms of it's work done.
2.) The amount of heat added to change the temperature is equivalent to the amount of work done by the gas. (ideally)
362 (Isobaric reactions)
1.)1000Pa*250m^3 = 250,000 J
2.)6000 J= (Vf-100)m^3 *300Pa
20 = Vf-100
Vf = 120 m^3
363-364 (Isochoric reactions)
1.) If volume is constant, as you change the temperature, the pressure changes, but no work is done.
2.) What changes is the internal energy.
364
1.) 0J Simpson
2.) 0J was innocent
365-366 (Isothermal reaction)
1.) In an isothermal reaction, the work done is equivalent to the amount of heat added to a system.
2.) W= nRTln(Vf/Vi)
367
1.) 1mol*(30+273)K*8.314 J/molK * ln(3/2) = 1,021 J
2.) .6mol * (25+273)K *8.314 J/molK * ln(3) = 1,633 J

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p358
1. Heat is the transfer of thermal energy. This energy is conserved.
2. Work is the transfer of mechanical energy. The first law of thermodynamics states that these energies, together, are conserved.
p359 internal energy
1. f-i 3000j-2000j= 1000J
2. 2500J-1700J= 800J
p360
1.  Force is equal to pressure
2. Work is dependent on volume and pressure for a gas.
p361
1. In an isobaric process pressure is constant.
2. For an isobaric process the volume of gas changes, but the weighted piston keeps pressure constant.
p362
1. (1000pa)(300m3-50m3)=250000J
2. w/p+vi 6000J/300pa+100m3=120m^3
p363-364
1. As gas increases pressure also increases while volume is constant.
2.  No work is done as volume stays the same
p364
1. no work V=0
2. no work v=0
p365-366
1. In an isothermal system temperature remains constant while everything else changes.
2. In an isothermal process internal energy doesn't change and heat remains the same as work.
p367
1. W=nRTIn(Vf/Vi)= (1.0)(8.31J/K)(273.15k+30 k) ln(3/2)=1020J
2. (.60)(8.31J/K)(273.15K+25K)(3/1)
1630J

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Page 358
1. The first law of Thermodynamics states that energy can neither be created nor destroyed.
2. Internal energy can change by doing work or a change in heat.
Page 359
1. 3,000J -2,000J=1,000J
2. 2,500 - 1,700J=800J
Page 360 and 361
1. Work depends  on volume and pressure for a gas.
2 .F= (pressure)(area)
3. An isobaric process involves work done by a gas without a change in pressure.
4. Pressure can remain constant in a piston if the weight on top is constant.
Page 362
1. Delta P=300 m^3 - 50 m^3
Delta P=250m^3
W=(1,000.0Pa)(250m^3)
W=250,000 J
2. 6,000.0J=(300Pa)(P2-100m^3)
20m^3=P2-100m^3
P=120 m^3
Page 363 and 364
1. An isochoric process: the pressure of a gas increasing but not the volume.
2. No work means V=0
Page 364
1. No work, as Delta V=0
2. No work, since Delta V=0
Page 365 and 366
1. In an isothermal process, the temperature is always constant
2. The product of pressure and volume remains constant.
Page 367
1. W=(1.0mol)(8.31J/K)(303.15K)ln(3.0/2.0)
W=1020J
2. W=(0.60mol)(298.15K)(8.13J/K)ln(3.0)
W=1,630J

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358
1. First Law of Thermodynamics deals with conservation of energy. -internal energy persists in situations of motion of atoms and molecules.
2. Work is the transfer of mechanical energy-for work in a gas, it must be compressed. Mechanical energy-work with systems where no energy lost in heat.
359
1. Change of U- 3000J-2000J=1000J
2. 2500J-1700J=800J
360
1.

# data[].attributes.body
pg.358
1)The first law of thermodynamics deals with energy conservation.
2) These three quantities-heat, work, and internal energy.
pg.359
1) 3,000 J of heat
2,000 J of work
3,000-2,000= 1000 J
2) 2,500 J of heat
1,700 J of work
2500-1700= 800 J
pg.360
1) Force= Pressure*area* distance
W= P*(triangle)V

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Pg.358
1)For mechanincal energy to be conserved, the system can not have energy lost to heat.
2)Heat, work and internal energy make up all of our energy.
Pg.359
1)3,000J-2,000 J= 1,000J
2)2500J-1700J= 800J

Pg.360
1)A gas only performs work if the gas expands
2)F= Pressure*area

Pg.361
1)Isobaric: process where the pressure stays constant
2) W=P delta V

Pg.362
1)(1,000)(300m^3-50m^3)=250000J
2)

Pg.363-364
1)isochoric: everything else changes as volume stays constant
2) If there is no work then V=0 since nothing is changing

Pg.364
1)No work done: V=0
2)No change in volume

Pg.365-366
1)In a isothermal system, temperature remains the same as everything else changes
2)P=nRT/V

Pg.367
1)
2)

# data[].attributes.body
pg.358
1. Uf-Ui=deltaU=Q-W
2. Heat, work, and internal energy make up the energy in a system.

# data[].attributes.body
Pg. 358
1) Heat, work, and the internal energy make up all of our energy
2) A system can lose energy by doing work on their surroundings.

Pg. 359 - Practice Questions
1)  3000J - 2000J = 1000J
2) 2500J - 1700J = 800J

Pg 360
1) W = P*(triangle)V
2) Graphing pressure versus volume graph compares them and the area under gives you work.

pg 361
1) With constant pressure, temperature change can change volume. Isobaric
2) In a graph, this relationship forms a rectangle.

Pg 362
1) W = (1000 Pa)(300 m^3 - 50 m^3) = 250000 J = 250 kJ
2) 6000 J = (300 Pa)(Vfin - 100)
20 = Vfin - 100; Vfin = 120 m^3

Pg 363
1) With constant volume, temperature can change pressure. Isochoric
2) Since volume stays the same, work done on the system equals zero.

Pg 364
1) W = 0 b/c (chg)V = 0; isochoric process
2) W = 0 b/c (chg)V = 0; isochoric process

Pg 365-366
1) With constant temperature, pressure and volume change. Isothermal
2) P = nRT/V; In a P vs V graph, curved top. W = nRT[ln(Vf/Vi)]

Pg 367
1) W = (1mol)(8.31J/K)(273.15K + 30K)ln(3/2) = 1020 J
2) Q = W = (0.6mol)(8.31J/K)(273.15K + 25K)ln(3/1) = 1630 J

# data[].attributes.body
P358
-The initial internal energy in a system, Ui, changes to a final internal energy, Uf, when heat, Q, is absorbed or released by the system and the system does work, -W, on its surroundings (or surroundings do work on the system) Uf-Ui=Change U= Q-W
- The quantity Q (Heat transfer) is positive when the system absorbs heat and negative when the system releases heat. The quantity W (work) is positive when the system does work on its surroundings and negative when the surroundings do work on the system.

P359
1) ΔU = 3000J-2000J= 1000J
2)ΔU = 2500J - 1700J = 800J
P360
Work = Force*time = Fs
Work = Pressure*Area*time = PAs
Work = PΔV =Pressure*Change in Velocity

P361
isobaric = constant pressure
W= PΔV
Work increases with an increase in volume when pressure is constant

P362
1.) W = (1000.0 Pa)(300.0m^3-50m^3) = 250,000J
2.) W = (6000.0J/300.0Pa) = 100.0m^3 = 120m^3

P363-364
- Volume is constant in an isochoric process, no work is done.
-The volume is constant, so Fs (force times distance) equals 0. No work is being done- the area under the graph is 0.

P364
1.
W=PΔV
P= 3000
ΔV= 0
W=3000*0=0
2.
ΔV=0
P=6000
W=6000*0=0

P365-6
- In an isothermal system, the temperature remains constant as other quantities change.
- P=(nRT)/V
- W=nRTln(Vf/vi)
- ΔU= Q-W
0=Q-W
Q=W

P367
1.) (1.0mol)(8.31 J/k)(273.15K+30.0K)(ln(3.0m^3)/(2.0m^3))
= 1,020J
2.) (.60mol)(8.31J/k)(273.15K+25.0K)(ln(3.0m^3)/(1.0m^3)) =1,630J

# data[].attributes.body
Pg. 358
1. The first law of Thermodynamics states that energy is neither created nor destroyed.
2. Internal energy can change by doing work or a change in heat.

Pg. 359
1. 3,000J -2,000J=1,000J
2. 2,500 - 1,700J=800J

Pg. 360-361
1. Work is dependent on volume and pressure for a gas.
2. You can find the change in volume through the area and displacement of the piston.
3. An isobaric process involves work done by a gas without a change in pressure.
4. Pressure can remain constant in a piston if the weight on top is constant.

Pg. 362
1. Delta P=300 m^3 - 50 m^3
Delta P=250m^3
W=(1,000.0Pa)(250m^3)
W=250,000 J
2. 6,000.0J=(300Pa)(P2-100m^3)
20m^3=P2-100m^3
P=120 m^3

Pg. 363-364
1. An isochoric process involves the pressure of a gas increasing, but not the volume.
2. This means there is no work, as volume does not change, and delta V=0

Pg. 364
1. No work, as Delta V=0
2. No work, since Delta V=0

Pg. 365-366
1. In an isothermal process, the temperature is always constant
2. The product of pressure and volume remains constant.

Pg. 367
1. W=(1.0mol)(8.31J/K)(303.15K)ln(3.0/2.0)
W=1020J
2. W=(0.60mol)(298.15K)(8.13J/K)ln(3.0)
W=1,630J