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.
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)
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.
Saibabu said:
1 decade ago
Here volume is constant therefore dv=0.
dw = pdv.
dw = 0.
According to first law dQ = du+dw.
dQ = dU.
Therefore if we are giving heat to the system then interna energy will also increase.
dw = pdv.
dw = 0.
According to first law dQ = du+dw.
dQ = dU.
Therefore if we are giving heat to the system then interna energy will also increase.
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.
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)
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.
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.
Dr. Aayush Gupta and Dr. Rajkiran Shrivastav said:
1 decade ago
Since we have considered a reversible process and closed system, in which work done = pdv, but at constant volume pdv work done is zero.
By first law of thermodynamics we have,
q = du + w,
therefore q=du.
By first law of thermodynamics we have,
q = du + w,
therefore q=du.
DIMPU SAI KRISHNAN said:
1 decade ago
In constant volume processes, pressure is changing so temperature is also changing if temperature is changing internal energy also change because internal energy is a function of temperature only.
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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