Mechanical Engineering - Thermodynamics - Discussion
Discussion Forum : Thermodynamics - Section 1 (Q.No. 9)
9.
When the gas is heated at constant volume, the heat supplied increases the internal energy of the gas.
Discussion:
27 comments Page 1 of 3.
Prashant mohadikar said:
7 years ago
As the increase in heat and change in temperature at no change in volume, means there is no any work done. Therefore this supplied heat energy not use and will be present in the gas molecules as its internal energy. Thank you.
(1)
Nagesh said:
9 years ago
According to Maxwell's equation,
dU = TdS - PdV.
So at the constant volume dV = 0.
dU = TdS - PdV.
So at the constant volume dV = 0.
(1)
Manish said:
9 years ago
According to the first law of thermodynamics when the system undergoing a process, dq = du + W.
As we know work transfer is zero in constant volume process.
From above equation if W = 0, Implies Heat supplied = Internal Energy.
So, as you increase the heat supply proportionally internal energy also increases.
As we know work transfer is zero in constant volume process.
From above equation if W = 0, Implies Heat supplied = Internal Energy.
So, as you increase the heat supply proportionally internal energy also increases.
(1)
Dheeraj said:
10 years ago
dQ = dw+du;
Constant vol: w = 0 then,
dQ = du.
Constant vol: w = 0 then,
dQ = du.
Saurabh said:
10 years ago
If we burn anything in specific volume the temperature will be increased and heat supplied will also be.
Sourabh thakur said:
10 years ago
Let, for ex, If we consider piston cylinder arrangement under constant volume and if energy is supplied to it, volume should remain constant.
So all the energy is used in increasing its internal energy and boundary work will be zero. That is the reason why we take W=0 in isochoric process.
So all the energy is used in increasing its internal energy and boundary work will be zero. That is the reason why we take W=0 in isochoric process.
Om prakash singh said:
10 years ago
At constant volume, w = 0.
And we know that Q = U+v (from first law of thermodynamics: DQ = du + dw).
From that relation we can see that Q = U+0.
i.e, Q = U
So, heat supplied increases the internal energy of the gas.
And we know that Q = U+v (from first law of thermodynamics: DQ = du + dw).
From that relation we can see that Q = U+0.
i.e, Q = U
So, heat supplied increases the internal energy of the gas.
Kunwar singh said:
1 decade ago
If we heat at a constant volume then W=O.
And according to thermodynamics first law:
dQ = dU+dW.
dQ = dU.
Then it heat supplied will be increase then it internal energy automatically increase because it final temperature will be increase.
And according to thermodynamics first law:
dQ = dU+dW.
dQ = dU.
Then it heat supplied will be increase then it internal energy automatically increase because it final temperature will be increase.
ROHIT KUMAR said:
1 decade ago
We know that in constant volume process volumes remain constant. So work done = 0.
Now,
DQ = dU+DW.
DW = 0.
So DQ = dU.
So heat supplied in constant volume process is used to increase it's internal energy.
Now,
DQ = dU+DW.
DW = 0.
So DQ = dU.
So heat supplied in constant volume process is used to increase it's internal energy.
Anil said:
1 decade ago
No pressure change means no work. In this case heat supplied is =internal energy. From 1st law.
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