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 2 of 3.
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:
1 decade 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.
Amar Dev said:
1 decade ago
Heat supplied = Work done + Change in internal energy.
dQ = W + dU.
dQ = PdV + dU [dV(change in volume) = 0 in case of constant volume].
So dQ = dU.
dQ = W + dU.
dQ = PdV + dU [dV(change in volume) = 0 in case of constant volume].
So dQ = dU.
Sanjeet kumar said:
1 decade ago
As we know that temperature is a function of inernal energy. i.e T= f(U)
Uttam kumar said:
1 decade ago
I can't understand que. In case of inter energy, they get increase but how the supplied energy get increase?
Hare ram said:
1 decade ago
In constant volume processes the volume remains const. The internal energy=m*cv* (change in temp) , the final temp is raising automatically internal energy increases.
Purusottam said:
1 decade ago
In constant volume process change in volume is zero. So boundary work Pdv is zero. from first law,
dq=du+dw, dw=0.
So dq=du, hence heat supply increases internal energy.
dq=du+dw, dw=0.
So dq=du, hence heat supply increases internal energy.
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