Mechanical Engineering - Thermodynamics - Discussion
Discussion Forum : Thermodynamics - Section 1 (Q.No. 6)
6.
An adiabatic process is one in which
Discussion:
49 comments Page 3 of 5.
AnKit said:
8 years ago
In Option B, Temperature of system changes with heat addition or Rejection. How could it possible?
I think the Option A is correct.
I think the Option A is correct.
Ramana murthy said:
8 years ago
Here Q=U+w it depends upon the piston moves from bdc to tdc.
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.
Manjeet Kumar said:
8 years ago
The internal energy of a ideal gas is function of temperature only. So, if internal energy will change adiabatically then temperature will also change.
U= Function of T only for ideal gas.
U= Function of T only for ideal gas.
Naeem said:
7 years ago
Temperature will change only in form of work done by system (decrease in temp) , or work done on system (increase in temp).
Pawan said:
1 decade ago
In adiabatic process q=0.
Then all the above conditions satisfied.
Then all the above conditions satisfied.
Abhishek said:
1 decade ago
Adiabatic process no heat enter and leave system.
So dQ = 0.
Now dQ = dU+dW.
Hence dW = -dU.
So dQ = 0.
Now dQ = dU+dW.
Hence dW = -dU.
Kshitiz said:
1 decade ago
If temperature of the gas changes then there will be some heat change or transfer, in that case how process will be adiabatic?
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.
Bipin said:
1 decade ago
An adiabatic process is constant heat process i.e. not heat transfer will occurs, if there is no heat transfer then how the temperature will change, therefor the answer is wrong as given above.
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