Download Test Thermodynamics Solutions

Solutions Thermodynamics Test Term 2, 2015
Define:
i.
System: the object(s) or physical situation that is being studied
Boundary: is what separates the system from the environment
Environment: everything external to the system that has bearing on the system
ii.
Thermodynamic process
A process that changes from one state (p,v, and t) to another state examples include: isobaric,
isovolumetric, isothermal, adiabatic, and polytrophic.
iii.
Adiabatic
A thermodynamic process where no heat is exchanged between the system and the
environment, ΔQ = 0
iv.
Polytrophic
Thermodynamic process where the relationship between pressure, volume is 𝑝𝑣 𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
v.
The four laws of thermodynamics
0th Law of thermodynamics: There exists a useful scalar quantity called temperature that is a
property of all thermodynamics systems. When 2 bodies (A and B) are in thermal equilibrium
(same temperature) with a third body (C), then all bodies (AC, BC, AB) are in equilibrium with
each other.
1st Law of thermodynamics: Energy is neither created or destroyed, energy can only change form.
𝑑𝑈
𝑑𝑄
There is exists a useful quantity called internal energy. 𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊 𝑜𝑟 ∫ 𝑑𝑡 = ∫ 𝑑𝑡 − ∫
𝑑𝑊
𝑑𝑡
2nd Law of thermodynamics: Heat must flow from hot to cold. The second law deals with entropy
or the measure of the quality of energy in natural processes and states if a process can happen in
nature.
3rd Law of thermodynamics: it is impossible by any procedure, no matter how idealized, to reduce
any system to absolute zero temperature in finite number of operations.
List as many as you know:
i.
System variables
Pressure, Volume, Temperature, Mass/Moles and Internal Energy
ii.
Equations of state
∆𝑈 = ∆𝑄 − ∆𝑊
𝑝𝑣 = 𝑛𝑅𝑇
𝑝𝑣 𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑎
) (𝑣 − 𝑏) = 𝑅𝑇
𝑣2
𝑝𝑉𝑚
𝑍=
𝑅𝑇
(𝑝 +
iii.
Thermodynamic processes
A process that involves heat transfer, work or a change in internal energy of a system.
1. Sketch the following processes on a pV diagram, Isobaric, isovolumetric, isothermal and
adiabatic
Pressure
A
Process A to B: Isothermal
Process A to C: Adiabatic
Process A to D: Isovolumetric
B
D
Process B to D: Isobaric
C
Volume
2. A gas expands from an initial state where p1 = 400 kPa and V1 = 0.2 m3 to a final state where p2 =
200 kPa. The relationship between pressure and volume during the process is pV = constant.
Sketch the process on a p-V diagram and determine the work, in kJ.
Pressure
p1V1
p2V2
Volume
𝑝𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑝1 𝑉1 = 𝑝2 𝑉2
𝑝1 𝑉1
𝑉2 =
𝑝2
400 𝑘𝑃𝑎 × 0.2 𝑚3
= 𝑚3
200 𝑘𝑃𝑎
𝑽𝟐
400 × 0.2 𝑚3
= 𝟎. 𝟒 𝒎𝟑
200
𝑉 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑉
1
𝑊 = ∫ 𝑝 𝑑𝑉 = ∫𝑉 2
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ln
= 𝑝1 𝑉1 ln
𝑉2
𝑉1
𝑑𝑉
𝑉2
𝑉1
𝑾 = 400(0.02) ln
0.4
= 𝟓𝟓. 𝟓 𝑱𝒐𝒖𝒍𝒆𝒔
0.2
3. One thousand kg of natural gas at volume of 10 m3 and 255 K is stored in a tank. If the pressure,
p, specific volume, v and temperature, T, of the gas are related by the following expression
𝑝 = [(5.18 𝑥 10−3 ) {
𝑘𝑃𝑎 𝑚3
}
𝑘𝑔 𝐾
𝑇
]−
𝑚3
(𝑣 − 0.002668
)
𝑘𝑔
(8.91 𝑥 10−3 ) 𝑘𝑃𝑎 𝑚9
{
}
𝑣2
𝑘𝑔2
where v is in m3/kg, T is in K, and p is in bar, determine the pressure of the tank.
𝑝 = [(5.18 𝑥 10
−3 )
𝑘𝑃𝑎 𝑚3
{
}
𝑘𝑔 𝐾
𝑝 = [(5.18 𝑥 10−3 )
𝑝 = [(5.18 𝑥 10−3 )
𝑉
𝑚
𝑣=
10 𝑚3
𝑚3
= 0.01
1000 𝑘𝑔
𝑘𝑔
(8.91 𝑥 10−3 ) 𝑘𝑃𝑎 𝑚9
]−
2 { 𝑘𝑔2 }
𝑚3
𝑚3
𝑚3
(0.01
− 0.002668
)
(0.01
)
𝑘𝑔
𝑘𝑔
𝑘𝑔
255 𝐾
(8.91 𝑥 10−3 ) 𝑘𝑃𝑎 𝑚9
255
𝑘𝑃𝑎
]{
}−
2 { 𝑘𝑔2 }
(0.01 − 0.002668 )
𝑚3
(0.01
)
𝑘𝑔
(8.91 𝑥 10−3 ) 𝑘𝑃𝑎 𝑚9
255
𝑘𝑃𝑎
]{
}−
{
}
𝑚9
(0.01 − 0.002668 )
𝑘𝑔2
(0.0001
)
𝑘𝑔2
𝑝 = [(5.18 𝑥 10−3 )
(8.91 𝑥 10−3 ) 𝑘𝑃𝑎
255
𝑘𝑃𝑎
]{
}−
{
}
(0.01 − 0.002668 )
(0.0001 )
𝑝= [
𝒑= [
𝑣=
(89.1 ) 𝑘𝑃𝑎
1.32
𝑘𝑃𝑎
]{
}−
{
}
(0.007332 )
180.0
]{
𝑘𝑃𝑎
}−
(89.1 ) 𝑘𝑃𝑎
{
} = 𝟗𝟏. 𝟏 𝒌𝑷𝒂
4. One mole of a monatomic gas goes in a cycle ABCA as in figure 3. Let (P0, V0, T0) be the
parameters at A and 2T0 the temperature at B. Solve algebraically:
i.
The values of TC and PB
ii.
The work done in each part: AB, BC and CA
iii.
The heat input in each part: AB, BC and CA
Given:
State
A
B
C
Pressure
P0
P0
Volume
V0
V0
2V0
Temperature
T0
2T0
Find: PB
𝑝𝑉 = 𝑛𝑅𝑇
Ideal Gas Law allows us to write
𝑃0 𝑉0 = 𝑃𝐵 𝑉𝐵 (1)
Also, at A
𝑃0 𝑉0 = 𝑛𝑅𝑇0 (2)
From the table above at B
𝑃𝐵 𝑉0 = 𝑛𝑅(2𝑇0 ) (3)
Thus, (2) ÷ (3) yields
𝑃0 𝑉0
𝑛𝑅𝑇0
=
𝑃𝐵 𝑉0 𝑛𝑅(2𝑇0 )
Canceling
𝑃0 1
= 𝑜𝑟 𝑷𝑩 = 𝟐𝑷𝟎
𝑃𝐵 2
Given
State
A
B
C
Pressure
P0
𝟐𝑷𝟎
P0
Volume
V0
V0
2V0
Temperature
T0
2T0
Find: TC
𝑝𝑉 = 𝑛𝑅𝑇
Ideal Gas Law allows us to write
𝑃0 𝑉0 = 𝑃𝐶 𝑉𝐶 (1)
Also, at A
𝑃0 𝑉0 = 𝑛𝑅𝑇0 (2)
From the table above at C
𝑃0 (2𝑉0 ) = 𝑛𝑅𝑇𝐶 (3)
Thus, (2) ÷ (3) yields
𝑃0 𝑉0
𝑛𝑅𝑇0
=
𝑃0 2𝑉0 𝑛𝑅𝑇𝐶
Canceling
1 𝑇0
=
𝑜𝑟 𝑻𝑪 = 𝟐𝑻𝟎
2 𝑇𝐶
State
A
B
C
Pressure
P0
𝟐𝑷𝟎
P0
Volume
V0
V0
2V0
Temperature
T0
2T0
𝟐𝑻𝟎
𝟐𝑷𝟎
𝟐𝑻𝟎
𝟐𝑻𝟎
Find the work done and heat input in each process
Process A -> B
∆𝑈 = ∆𝑄 − ∆𝑊
𝑉𝐵
𝑊 = ∫ 𝑝 𝑑𝑉 = ∫ 𝑝 𝑑𝑉
𝑉𝐴
Inputting values
𝑽𝟎
𝑾 = ∫ 𝒑 𝒅𝑽 = 𝟎
𝑽𝟎
∆𝑈 =
3𝑛𝑅(𝑇𝐵 − 𝑇𝐴 )
3(1)𝑅(2𝑇0 − 𝑇0 )
=
2
2
3𝑅𝑇0
∆𝑈 =
2
Plugging into Energy equation
3𝑅𝑇0
= ∆𝑄 − 0
2
or
∆𝑸 =
𝟑𝑹𝑻𝟎
𝟐
State
A
B
C
Pressure
P0
𝟐𝑷𝟎
P0
Volume
V0
V0
2V0
Temperature
T0
2T0
𝟐𝑻𝟎
𝟐𝑷𝟎
𝟐𝑻𝟎
𝟐𝑻𝟎
Process B -> C
∆𝑈 = ∆𝑄 − ∆𝑊
𝑉𝐶
𝑊 = ∫ 𝑝 𝑑𝑉 = ∫ 𝑝 𝑑𝑉
𝑉𝐵
Because pV = constant
𝑉𝐶
𝑊= ∫
𝑉𝐵
𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑑𝑉
𝑉
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ln
𝑉𝐶
𝑉𝐵
Because 𝑝𝐵 𝑉𝐵 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑉𝐶
= 𝑝𝐵 𝑉𝐵 ln
𝑉𝐵
Putting in values for pB, VB and VC
2𝑉0
= 2𝑝0 𝑉0 ln
𝑉0
𝑾 = 𝟐𝒑𝟎 𝑽𝟎 𝐥𝐧 𝟐
∆𝑈 =
3𝑛𝑅(𝑇𝐶 − 𝑇𝐵 )
3(1)𝑅(2𝑇0 − 2𝑇0 )
=
2
2
∆𝑼 = 𝟎
Plugging into Energy equation
0 = ∆𝑄 − 2𝑝0 𝑉0 ln 2
or
∆𝑸 = 𝟐𝒑𝟎 𝑽𝟎 𝐥𝐧 𝟐
State
A
B
C
Pressure
P0
𝟐𝑷𝟎
P0
Volume
V0
V0
2V0
Temperature
T0
2T0
𝟐𝑻𝟎
𝟐𝑷𝟎
𝟐𝑻𝟎
𝟐𝑻𝟎
Process C -> A
∆𝑈 = ∆𝑄 − ∆𝑊
𝑊 = ∫ 𝑝 𝑑𝑉
Because pressure is constant C -> A
𝑉𝐴
𝑊 = 𝑝 ∫ 𝑑𝑉 = ∫
𝑑𝑉
𝑉𝐶
= 𝑝(𝑉𝐴 − 𝑉𝐶 )
= 𝑝0 (𝑉0 − 2𝑉0 )
𝑾 = −𝒑𝟎 𝑽𝟎
∆𝑈 =
3𝑛𝑅(𝑇𝐴 − 𝑇𝐶 )
3(1)𝑅(𝑇0 − 2𝑇0 )
=
2
2
3𝑅𝑇0
∆𝑈 = −
2
Plugging into Energy equation
−
3𝑅𝑇0
= ∆𝑄 − (−)𝑝0 𝑉0
2
or
∆𝑸 = −
𝟑𝑹𝑻𝟎
− 𝒑𝟎 𝑽𝟎
𝟐