Work done in Isothermal and adiabatic Process
-
Upload
deepanshu-chowdhary -
Category
Education
-
view
382 -
download
0
Transcript of Work done in Isothermal and adiabatic Process
Work Done in Isothermal Work Done in Isothermal And Adiabatic ProcessAnd Adiabatic Process
From: DEEPANSHU CHOWDHARYFrom: DEEPANSHU CHOWDHARYRoll no: 05Roll no: 05
Class: 11Class: 11thth A A
Isothermal processIsothermal process
• P,V may change but temperature is P,V may change but temperature is constant.constant.
• The cylinder must have conducting wallsThe cylinder must have conducting walls• It must happen very slowly so that heat It must happen very slowly so that heat
produced during compression is absorbed produced during compression is absorbed by surroundings and heat lost during by surroundings and heat lost during compression is supplied by surroundings. compression is supplied by surroundings.
Adiabatic processAdiabatic process
• In an adiabatic process, the system is insulated from the In an adiabatic process, the system is insulated from the surroundings and heat absorbed or released is zero. Since surroundings and heat absorbed or released is zero. Since there is no heat exchange with the surroundings,there is no heat exchange with the surroundings,
• When expansion happens temperature falls When expansion happens temperature falls • When gas is compressed, temperature rises.When gas is compressed, temperature rises.
Kinds of ProcessesKinds of Processes
Often, something is held constant. Examples:
dV = 0 isochoric or isovolumic processdQ = 0 adiabatic processdP = 0 isobaric processdT = 0 isothermal process
• Work done when PV = nRT = constant Work done when PV = nRT = constant P = nRT / V P = nRT / VIsothermal processesIsothermal processes
final
initial
)curveunder area( dVpW
f
i
f
i
V
V
V
V
/ nRT/ nRT VdVVdVW
)/VV( nRT ifnW
p
V
3 T1
T2T3T4
Adiabatic Adiabatic ProcessesProcesses An adiabatic process is process in which there is no
thermal energy transfer to or from a system (Q = 0)
A reversible adiabatic process involves a “worked” expansion in which we can return all of the energy transferred.
In this casePV = const.
All real processes are not.
p
V
2
13
4T1
T2
T3T4
8
Adiabatic ProcessAdiabatic Process
•For an ideal gas, and most real For an ideal gas, and most real gasses,gasses,•đQ = dU + PdV đQ = dU + PdV •đQ = CđQ = CVVdT + PdV,dT + PdV,..
•Then, when Then, when đQđQ = 0, = 0, VCPdVdT
9
Adiabatic ProcessAdiabatic Process
nRVdPPdV
CPdV
nRVdPPdVdT
nRPVT
V
Then,
and ,
For an ideal gas, PV=nRT, so
10
Adiabatic ProcessAdiabatic Process
nRVdP
nRCCnRPdV
nRVdP
nRCPdV
nRVdPPdV
CPdV
nRVdPPdVdT
nRpVT
V
V
VV
0
110
Then,
and ,
11
Adiabatic ProcessAdiabatic Process
V
p
V
P
PV
V
V
CC
VdPPdVVdPCCPdV
CCnR
VdPC
CnRPdV
where,
0
0
12
Adiabatic ProcessAdiabatic Process
constant
constantlnlnln
constantlnln
,integrated becan which ,0
0
PV
PVPV
PVP
dPVdV
VdPPdV
13
Adiabatic ProcessAdiabatic Process
constant
constant
as, expressed be alsocan this of help With the
constant
1
1
PT
TV
nRTPVPV
14
for “Ideal Gasses”for “Ideal Gasses”
33.1621 :polyatomic
40.1521 :diatomic
67.1321 :monatomic
21
Combinations of Isothermal & Combinations of Isothermal & Adiabatic Adiabatic ProcessesProcessesAll engines employ a thermodynamic cycleW = ± (area under each pV curve)Wcycle = area shaded in turquoise
Watch sign of the work!
ISOTHERMAL PROCESS: ISOTHERMAL PROCESS: CONST. TEMPERATURE, CONST. TEMPERATURE, T = 0, T = 0, U U
= 0= 0
NET HEAT INPUT = WORK OUTPUTNET HEAT INPUT = WORK OUTPUT
Q = Q = U + U + W ANDW ANDQ Q = = WW
U = 0
U = 0
QQOUTOUT
Work Work InIn
Work Work OutOut
QQININ
WORK INPUT = NET HEAT OUTWORK INPUT = NET HEAT OUT
ISOTHERMAL EXAMPLE ISOTHERMAL EXAMPLE (Constant (Constant T):T):
PAVA =
PBVB
Slow compression at constant temperature: ----- No change in UNo change in U.
U = U = TT = = 00
B
APA
V2 V1
PB
ISOTHERMAL EXPANSION (ISOTHERMAL EXPANSION (Constant Constant T)T)::
400 J of energy is absorbed by gas as 400 J of work is done on gas.
T = U = 0
U = T = 0
BBAA
PA
VA VB
PB
PAVA = PBVB
TA = TB
ln B
A
VW nRTV
Isothermal Work
Q = Q = U + U + W ; W ; W = -W = -U or U or U = -U = -WW
ADIABATIC PROCESS: ADIABATIC PROCESS: NO HEAT EXCHANGE, NO HEAT EXCHANGE, Q = 0Q = 0
Work done at EXPENSE of internal energy INPUT Work INCREASES internal energy
Work Out
Work InU +U
Q = 0
W = -U U = -W
ADIABATIC EXAMPLE:ADIABATIC EXAMPLE:
Insulated Walls: Q =
0
BA
PPAA
VV11 V V22
PPBB
Expanding gas Expanding gas does work with does work with zero heat loss. zero heat loss. Work = -Work = -UU
ADIABATIC EXPANSION:ADIABATIC EXPANSION:
400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat exchange is ZERO. Q = 0Q = 0
Q = 0
B
APPAA
VVAA VVBB
PPBB
PPAAVVA A PPBBVVBB
TTA A TT B B
=
A A B BP V P V
ADIABATIC EXAMPLE:
Q = 0
AA
BBPPBB
VVBB V VAA
PPAA PAVA PBVB
TTA A TT B B
=
PPAAVVAA = P = PBBVVBB
Example 2: A diatomic gas at 300 K and 1 atm is compressed adiabatically, decreasing its volume by 1/12. (VA = 12VB). What is the new pressure and temperature? ( = 1.4)
ADIABATIC (Cont.): FIND PADIABATIC (Cont.): FIND PBB
Q = 0
PB = 32.4 atm or 3284 kPa
1.412 B
B AB
VP PV
1.4(1 atm)(12)BP
PPAAVVAA = P = PBBVVBB
AA
BBPPBB
VVBB 12VVBB
1 1 atmatm
300 K Solve for Solve for PPBB::
AB A
B
VP PV
ADIABATIC (Cont.): FIND TADIABATIC (Cont.): FIND TBB
Q = 0
TB = 810 K
(1 atm)(12V(1 atm)(12VBB)) (32.4 atm)(1 V(32.4 atm)(1 VBB)) (300 K)(300 K) TT B B
==
AA
BB32.4 32.4 atmatm
VVBB 12 12VVBB
1 1 atmatm
300 K
Solve for Solve for TTBB
TTBB=?=?A A B B
A B
P V P VT T
ADIABATIC (Cont.): ADIABATIC (Cont.): If VIf VAA= 96 cm= 96 cm33 and Vand VAA= 8 cm= 8 cm33, FIND , FIND WW
Q = 0
W = - W = - U = - nCU = - nCVV TT & & CCVV== 21.1 j/mol 21.1 j/mol KK
AA
B32.4 32.4 atmatm
1 1 atmatm
300 K
810 KSince Since Q = Q =
0,0,
W = - W = - UU 8 cm8 cm3 3 96 cm96 cm3 3
Find n Find n from point from point
AAPV = nRTPV = nRT
PVPV
RTRT n =n =
ADIABATIC (Cont.): ADIABATIC (Cont.): If VIf VAA= 96 cm= 96 cm33 and Vand VAA= 8 cm= 8 cm33, FIND , FIND WW
AA
BB32.4 32.4 atmatm
1 atm
300 K
810 K
8 cm8 cm3 3 96 cm96 cm33
PVPV
RTRT n =n = = = (101,300 Pa)(8 x10(101,300 Pa)(8 x10-6-6 m m33))
(8.314 J/mol K)(300 K)(8.314 J/mol K)(300 K)nn = 0.000325 mol = 0.000325 mol & & CCVV= 21.1 j/mol K= 21.1 j/mol K
TT = 810 - 300 = 510 K = 810 - 300 = 510 K
W = - W = - U = - nCU = - nCVV TT
W = - 3.50 J
THANKS FOR
THANKS FOR
WATCHING!!!
WATCHING!!!