Mechanical Engineering - Thermodynamics - Discussion
Discussion Forum : Thermodynamics - Section 1 (Q.No. 6)
6.
An adiabatic process is one in which
Discussion:
49 comments Page 1 of 5.
Javed said:
10 years ago
An adiabatic process is one that occurs without transfer of heat or matter between a thermodynamic system and its surroundings; energy is transferred only as work. The adiabatic process provides a rigorous conceptual basis for the theory used to expound the first law of thermodynamics, and as such it is a key concept in thermodynamics.
Some chemical and physical processes occur so rapidly that they may be conveniently described by the "adiabatic approximation", meaning that there is not enough time for the transfer of energy as heat to take place to or from the system.
In way of example, the adiabatic flame temperature is an idealization that uses the "adiabatic approximation" so as to provide an upper limit calculation of temperatures produced by combustion of a fuel. The adiabatic flame temperature is the temperature that would be achieved by a flame if the process of combustion took place in the absence of heat loss to the surroundings.
Some chemical and physical processes occur so rapidly that they may be conveniently described by the "adiabatic approximation", meaning that there is not enough time for the transfer of energy as heat to take place to or from the system.
In way of example, the adiabatic flame temperature is an idealization that uses the "adiabatic approximation" so as to provide an upper limit calculation of temperatures produced by combustion of a fuel. The adiabatic flame temperature is the temperature that would be achieved by a flame if the process of combustion took place in the absence of heat loss to the surroundings.
Lokesh chauhan said:
7 years ago
For example, take an insulated piston cylinder device containing some arbitrary gas (common example in thermodynamics problems). If you were to push the piston down (thus compressing the contents of the cylinder), the temperature of the gas in the cylinder would increase. If the piston was pulled out (thus expanding the gas) the temperature would decrease. In either case, no heat would flow to or from the system, but the temperature would change approximately according to the relationship PV=nRT (for the most general cases).
So that's why this is the right answer.
So that's why this is the right answer.
(3)
Ankit Kumar Sharma said:
8 years ago
Option B is also correct because as we have studied in Air refrigeration system that due to throttling process (due to the friction occurs between container and gas atoms) the heat generated and is absorbed by a gas atom and hence a slight change occurs in the temperature of gas.
That's why the temperature changes occur in the adiabatic process also.
As all the three options are true for the question so the right answer is OPTION:- D.
That's why the temperature changes occur in the adiabatic process also.
As all the three options are true for the question so the right answer is OPTION:- D.
Amit said:
1 decade ago
Due to change in internal energy, the temperature may change.
U = (KE + PE + IE).
KE is the kinetic energy of molecules = 3/2 kT (k = Boltzmann's const. & T = temperature).
PE is the potential energy.
IE is the intramolecular energy which may also lead to a change in temperature due to molecular re orientations.
U = (KE + PE + IE).
KE is the kinetic energy of molecules = 3/2 kT (k = Boltzmann's const. & T = temperature).
PE is the potential energy.
IE is the intramolecular energy which may also lead to a change in temperature due to molecular re orientations.
Vasanth kumar said:
1 decade ago
As per adiabatic process assumption there will be no heat transfer in or out of the system. First condition is satisfied.
The internal energy within cause the work, So that temperature changes.
As per first law of thermodynamics dQ equals to du+dw.
Heat transfer is zero du equal -dw. So that third is also correct.
The internal energy within cause the work, So that temperature changes.
As per first law of thermodynamics dQ equals to du+dw.
Heat transfer is zero du equal -dw. So that third is also correct.
Prasad said:
1 decade ago
In adiabatic process, no heat transfer takes place. And there, may be a temperature change, and it need not involve nay heat transfer to it. It may occur naturally as well. For eg. I am compressing a cylinder, automatically pressure will rise, so also the temp. Same as reverse.
And dq = -dw.
So answer is correct.
And dq = -dw.
So answer is correct.
Saravanakumar said:
1 decade ago
@Muthamizh Selvan There is one.
Q = Del E + Del W.
Del E = dU+ dE, here dU is the heat produced due to work. So there is heat energy is induced inside the system, there the temperature will changes. Still there is no heat transfer across the boundary, Since the process is adiabatic.
Del W = Del W.
Q = Del E + Del W.
Del E = dU+ dE, here dU is the heat produced due to work. So there is heat energy is induced inside the system, there the temperature will changes. Still there is no heat transfer across the boundary, Since the process is adiabatic.
Del W = Del W.
Aarti Mirge said:
1 decade ago
As per second law of thermodynamics, dQ=TdS, in adiabatic process dS=0, so dQ=0.
And hence dU=-dW, i.e, work is done on account of internal energy and as internal energy is a function of temperature for an ideal gas so temp will change although there is no heat transfer in adiabatic process.
And hence dU=-dW, i.e, work is done on account of internal energy and as internal energy is a function of temperature for an ideal gas so temp will change although there is no heat transfer in adiabatic process.
Sandeep singh said:
1 decade ago
In adiabatic process n heat enters or leave, but if system works itself by spending its internal energy then definitely temperature will change, because system did some work but did not take heat from external sources. So it has to spend its internal energy if it is going to do some work.
Mohammad Khurram Saeed said:
1 decade ago
Second law dQ =TdS, dS = 0 for adiabatic process I.e isentropic and irreversible process.
(Constant temperature heat change as the process is quasi static leading to irreversible process).
Hence, dQ = 0.
dQ= dU + dW.
Hence, dU = - dW.
(Constant temperature heat change as the process is quasi static leading to irreversible process).
Hence, dQ = 0.
dQ= dU + dW.
Hence, dU = - dW.
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