# data[].attributes.body
p. 358
ΔU = Q - W (The change in the internal energy of a system is equal to the heat absorbed or released by the system, Q, minus the work done on or by the system, W.)
In any process, energy is conserved in the form of heat, work, and internal energy.

p. 359
1. ΔU = 3000 J - 2000 J = 1000 J
2. ΔU = 2500 J - 1700 J  = 800 J

p. 360
W = PΔV (The work done by a gas equals the pressure times the change in volume.)
In a pressure-volume volume graph, the curve shows how pressure and volume change in relation to each other, and the area under the curve shows how much work is done.

p. 361
In an isobaric process, pressure remains constant.
The work done in an isobaric process is the area under its graph, PΔV.

p. 362
1. W = (1000.0 Pa)(300.0 m³ - 50.0 m³) = 250000 J
2. ΔV = 6000.0 J / 300.0 Pa = 20 m³ = Vf - 100.0 m³
Vf = 120 m ³

p. 363-364
In an isochoric process, volume remains constant.
Because volume is constant, an isochoric process does not do work.

p. 364
1. W = PΔV, ΔV = 0, W = 0 J
2. W = PΔV, ΔV = 0, W = 0 J

p. 365-366
In an isothermal process, temperature remains constant.
Because n (moles of gas), R (the ideal gas constant) and T (temperature) in the ideal gas law PV = nRT remain constant, the product of pressure and volume is constant in an isothermal process.
The work done in an isothermal process is given by W = nRTln(Vf / Vi). Note that R = 8.31 J/mol·K.

p. 367
1. W = (1.0 mol)(8.31 J/mol·K)(303.15 K)ln(3.0 m3 / 2.0 m³) = 1020 J
2. W = (0.60 mol)(8.31 J/mol·K)(298.15 K)ln(3.0 m³ / 1.0 m³) = 1630 J

# data[].attributes.body
P. 363-364
- Even if pressure increases, the volume will stay the same.

P. 364
1.) No work is done = 0.
2.) No work is done = 0.

P. 365-366
- In an isothermal system, the temperature remains constant as other quantities change.
- As work changes the formula is P= nRT/V

P. 367
1.) (1M) (8.31 J/K) (273+30) Ln(3/2) = 1020.93 J
2.) (.6M) (8.31 J/K) (273+25) = 1632.35J

# data[].attributes.body
P.358
1.The first law of thermodynamics states that energy can't be created or destroyed.
2.The first law of thermodynamics ties three of the quantities of energy together
P.359
1.3000J-2000J = 1000J
2) 3500J-1700J = 800J
P.360
1.Gas performs work when it expands.
2. Work=Pressure*Change in volume
P.361
1.Isobaric process when the pressure does not change.
2. As the volume of the gas changes, the weight of the piston keeps the pressure constant.
P.362
1. 1000 Pa * (300-50) = 250, 000 J
2.(6000 J / 300 Pa) + 100 m^3 = 120 m ^3
P.363-364
1.A constant volume is called a isochoric process
2. Because the volume is constant, the work is not done.
P.364
1. No work = 0J
2. NO work= 0J
P365-366
1. In an isothermal system, the temperature remains constant while volume and pressure change.
2.W=nRT*ln(Vf/Vi)                       PV=nRT
P.367
1.1(8.31 J/K)(303.15 K) ln(3/2)= 1,020 J
2. 0.6(8.31 J/K)(298.15 K) ln(3/1)= 1,630 J

# data[].attributes.body
p358
1. First law of thermodynamics states that the initial internal energy in a system changes to a final internal energy when heat is absorbed or released by the system and the system does work on its surroundings.
2. Since the first law ties all of these together, if you know two you can find the third.
p359
1. U= 3000J-2000J=1000J
2.U=2500J-1700J=800J
p360
1. Gas performs work only if it expands
2. W= (Pressure)(Change in Volume)
p361
1. Isobaric is process where pressure stays constant.
2. Change in volume may occur but pressure still stays constant.
p362
1. W=(1000)(300-50)-250,000J
2.Vf= 6000/300 +100= 120m^3
p363-364
1. If volume is constant, its an isochoric process
2. Since volume is constant, no work is done.
p364
1. No change in volume->isochoric -> no work done -> 0J
2. No change in volume -> isochoric -> no work done -> 0J
p365-366
1. If temperature is the only thing constant in a system, then it is an isothermal system.
2. Area under the curbe of a volume vs pressure is the work done in an isothermal process
p367
1. W=(1))(8.31)(273.15+30) ln(3/2)=1020J
2. W=(.6)(8.31)(273.15+25)ln(3/2)=1630J

# data[].attributes.body
Pg. 358
1.) Initial internal energy changes into final internal energy when heat is absorbed by the system which causes the system not to work.
2.) The first law of thermodynamics states that energies that are formed together become conserved.
Pg. 359
1) 3000J-2000J = 1000J
2) 3500J-1700J = 800J
Pg. 360
1.) Gas expands then it performs
2.) Work is equal to pressure times the change in volume
Pg. 361
1.) When a process has a constant pressure, then it is isobaric.
2.) Yes, it is possible for the volume to change in the isobaric process.
Pg.362
1) 1000 Pa * (300-50) = 250, 000 J
2) (6000 J / 300 Pa) + 100 m^3 = 120 m ^3
Pg. 363-364
1.) A constant volume is called a isochoric process
2.) Since the volume is constant, the work is not done
Pg. 364
1.) V=0
2.) V=0
Pg. 365-366
1.) In an isothermal process the temperature remains constant
2.) PV= nRT is always the same
Pg. 367
1.  (1 M)(8.31 J/K)(273+30)ln(3/2) = 1020.93 J
2. (0.6 M)(8.31 J/K)(273+25)ln(3/1) = 1632.35 J

# data[].attributes.body
p 358
1) - The 1st law of thermodynamics states that energy cannot be created or destroyed.
- For mechanical energy to be conserved  you have to work with systems where no energy is lost to heat
p 359
1) 3000J-2000J = 1000J
2) 3500J-1700J = 800J
p 360
- if gas expands then it performs work
- work is equal to the pressure times the change in volume
p 361
- If a process has a constant pressure then it is isobaric
- it is possible for the volume to  change in the isobaric process
p 362
1) 1000 Pa * (300-50) = 250, 000 J
2) (6000 J / 300 Pa) + 100 m^3 = 120 m ^3
p 363 - 364
- A constant volume is called a isochoric process
- the volume is constant so work is not done
p 364
1) No work is done, v = 0
2) No work is done, v = 0
p 365 - 366
- In an isothermal process the temperature remains constant
- PV = nRT is always the same
p 367
1.  (1 M)(8.31 J/K)(273+30)ln(3/2) = 1020.93 J
2. (0.6 M)(8.31 J/K)(273+25)ln(3/1) = 1632.35 J

# data[].attributes.body
P. 358:
1) A system's initial internal energy changes to final internal energy when heat is absorbed or released by the system and the system does work.
2) The first law of thermodynamics ties three of the quantities of energy together
P. 359:
1) 3,000 J-2,000 J= 1,000 J
2) 2,500 J-1700 J= 800J
P. 360:
1) A gas performs work only when the gas expands.
2) The curve drawn in a pressure v. volume graph shows how pressure and volume change in relation to each other; the area under the curve shows how much work has been done.
P. 361:
1) In a process where pressure remains constant, the process is called isobaric.
2) An isobaric system has constant pressure but the volume of a system's gas will change.
P. 362:
1) 1,000 Pa (300 m^3- 50 m^3)= 250,000 J.
2) (6,000 J/ 300 Pa) +100 m^3= 120 m^3
P. 363-364:
1) An isochoric process is one where the volume is constant but the pressure is not.
2) No work is done in an isochoric process since the volume is constant.
P. 364:
1) 0 J; isochoric process so no work done.
2) 0 J; isochoric process so no work done.
P. 365-366:
1) In an isothermal system, the temperature remains constant while volume and pressure change.
2) The internal energy of an isothermal process does not change since the temperature remains constant and internal energy for an ideal gas equals (3/2)nRT.
P. 367:
1) 1(8.31 J/K)(303.15 K) ln(3/2)= 1,020 J
2) .6(8.31 J/K)(298.15 K) ln(3/1)= 1,630 J

# data[].attributes.body
P. 358
(2 Key Pts.)- The first law of thermodynamics deals with energy conservation. There are three forms of energy, Internal Energy, Heat, and Work.
-  The three quantities make up all the energy that needs to be considered. A system can lose its energy by doing work on its surroundings.
1.) 3,000 J - 2,000J = 1,000J
2.) 2500 J - 1700 J= 800J

P. 360
1.)  Gas performs the work needed only when the gas expands.
2.) Change is quasi-static, meaning that change comes slowly enough that the pressure and temperature are the same throughout the system's volume.

P. 361
1.) Isobaric is a process where the pressure stays constant.
2.) As the volume of the gas changes, the weight of the piston keeps the pressure constant.

P. 362
1.) 1000 pa * (300-50) = 250000 J
2.) (6000 J/ 300 pa) + 100 m^3= 120 m^3

P 363-364
- An isochoric process is when the volume is not constant.
-

# data[].attributes.body
p358
1.) One of the forms of energy involved with energy conservation is internal energy. Internal energy is energy that resides in the motion of the atoms and molecules (vibrations and random jostling). 2.)The first law of thermodynamics states that energies, together, are conserved.
p359
1.)
2.)
p360
1.) The ways in which these quantities (volume, pressure, temperature,  etc) vary as work is done determine the final state of the system.
2.) A gas performs work only if the gas expands.
p361
1.) It is an isobaric process when the pressure stays constant.
2.) "-baric" means pressure
p362
1.)
2.)
p363-364
1.) An isochoric process is where the volume stays constant.
2.) Even if the pressure increases, the volume will still stay the same.
p364
p365-366
1.) In an isothermal system, the temperature remains constant as other quantities change.
2.) keep the volume constant
p367

# data[].attributes.body
Pg 358
- heat, work,  and internal energy make up all the energy in a system
- Uf - Ui = change in U = Q - W
where U is internal energy, Q is heat, and W is work
Pg 359
1. 3000J - 2000J= 1000J
2. 2500J - 1700J = 800J
Pg 360
- gas performs work only if gas expands
- work= pressure*change in volume
Pg 361

- in a system where pressure remains constant it is called isobaric

- the volume may change in an isobaric system
Pg 362
1. 1000Pa * (300 - 50)= 250000J
2. 6000J = 300Pa * (100- x)
x= 120m^3
Pg 363- 364
- a system where volume remains constant is called isochoric
- since volume is constant in an isochoric process, no work is done
1. W= 0 J because the volume is constant
2. W = 0 J because volume is constant
Pg 365- 366
- in an isothermal system the temperature remains constant
- the work done by an isothermal system is given by
W= nRT*ln(Vf/Vi)
Pg 367
1. W = 1*3.81J/K*(273.15K+30K) * ln(3/2) = 1020J
2. W = 0.6*8.31J/K*(273.15K+25K)* ln(3/1) = 1630J

# data[].attributes.body
p. 358
- In the First Law of Thermodynamics, it states that energy must be never destroyed or created.
- The first law of thermodynamic states that all of the quantities are connected with each other.

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

p. 360
- The change in area times the distance is the change in volume.
- Force is equivalent to the to the pressure times areas.

p. 361
- Isobaric process when the pressure does not change.
- The suffix -baric means pressure.

p. 362
1. 1000PA(250 M^3)= 250000 J
2. 6000/300 + 100 m^3 = 120 m^3

p. 363-364
-  If the volume is constant, the system is called isochoric
-  Because the volume is constant, no work can be done.

p. 364
1. 0 Joules because no change in volume.
2. 0 Joules because no change in volume.

p. 365-366
- An isothermal system is when the temperature remains constant
- The formula for work done in this system W=nRT*ln(Vf/Vi)

p. 367
1. (8.31 J/K)(303.15 K) ln(3/2) = 1020 J
2. .6(8.31 J/K)(298.15 K) ln(3) = 1633.171

# data[].attributes.body
p358
The first law of thermodynamics states that energy cannot be created or destroyed.
- The first law of thermodynamics ties all the quantities together
1.  3,000 J-  2000 J= 1000J
2. 2500 J - 1700 J= 800 J
p360
- A gas performs work only if thr gas expands. W=Fs
- Force equals pressure P times area, so you can write work as pressure times area times distance. W= PAs
p361
- When you have a process where the pressure stays constant, its called isobaric
- Work is the area under the curve of Pressure/Volume
p362
1. (1000.0 pascals) (300m^3- 50m^3)= 250000J
2.( 6000J/ 300 Pascals)+ 100m^3= 120m^3
p363-364
1. When a volume is constant, it is isochoric
2. As the gas inside the spray can heats up, its pressure increases, nut its volume stays the same.
364
1. Still at the same volume, so change in volume is zero, meaning work is zero
2. Volume is constant, so work is zero
p365-366
-In an isothermal systems, the temperature remains constant as other quantities change
- The relation between P and V is P=(nRT)/V
p367
1.1. W = 1*3.81J/K*(273.15K+30K) * ln(3/2) = 1020J
2. 0.6(8.31J/K) (298.15K) ln(3)= 1633.171J

# data[].attributes.body
p358
1. First law of thermodynamics

# data[].attributes.body
1. The first law of thermodynamics deals with the energy conservation. Energy consists of three quantities - heat, work, and internal work.
2. 3000-2000 = 1000J / 2500-1700 = 800
3. The formula W = P (change in V) makes graphs of pressure versus volume very useful in thermo-dynamics.
4. Isobaric refers to the process where pressure stays constant.
Work s the integral of the pressure and volume graph.
5. (1000)(250) = 250000 J
(200)(6000) = 1200000 J
6. If the volume is constant, then you have an isochoric process. As the gas inside the spray heats up, its pressure increases.
7. 2000(50) = 100000
8. In an isothermal system, the temperature remains constant. the relationship is P = (nRT)/V
1.  (1.0)(30)(1.0) = 30 J

# data[].attributes.body
p358
1. when a system does work on its surroundings and gives off no waste heat, its internal energy changes by the amount of work done by the system
2. the total energy of a system includes heat, work , and the internal energy of a system, which is in the motion of the vibrations of atoms and molecules

p359
1. 3000J - 2000J = 1000J
2. 2500J - 1700J = 800J

p360
1. A gas only performs work if it expands
2. W = F*d, or W = PV for gases

p361
1. when you have a process where the pressure stays constant, its called isobaric
2. work is area under the graph because W=PV

p362
1. (1000 pa)(300-50) = 250,000 J
2. 6000 = (300)(V_f - 50)
V_f = 70

p363-364
1. when volume is constant, you have an isochoric process
2. no work is done because the volume is constant

p364
1. V = 0, so no work done
2. V = 0, so no work done

p365-366
1. if temperature is constant, system is isothermal
2. everything else can change, but they have to adjust to keep the product of PV=nRT the same

p367
1.  (1 M)(8.31 J/K)(273+30)ln(3/2) = 1020.93 J
2. (0.6 M)(8.31 J/K)(273+25)ln(3/1) = 1632.35 J