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How to solve constants out of the internal energy equation?
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$begingroup$
Imagine we deal with a new kind of matter, whose state is described by:
$$PV = AT^3$$
Its internal energy is given by:
$$U = BT^n lnleft(frac{V}{V_0}right) + f(T)$$
Where $A, B$ and $V_0$ is a constant and $f(T)$ is a polynomial function.
Find B and n.
This is what I know:
The given expressions remind me of adiabatic compression/expansion. If we assume quasistatic adiabatic compression/expansion we know that heat won't get out/in the system.
$$Delta U = -W$$
And work is:
$$W = -PDelta V$$
Some thoughts on how to solve the problem
We notice here that we are dealing with a non-ideal gas. Assuming that the above equations are correct and using first thermodynamics law one gets:
$$mathrm{d}U = [nBT^{n-1} lnleft(frac{V}{V_0}right) + f'(T)]mathrm{d}T$$
$$[nBT^{n-1} lnleft(frac{V}{V_0}right) + f'(T)]mathrm{d}T = frac{AT^3}{V}mathrm{d}V$$
I do not see a solution to this differential equation.
There has to be a easier way to get both $B$ and $n$ but how?
Thanks.
thermodynamics equation-of-state
edited 2 hours ago
Gaurang Tandon
5,46262864
asked 3 hours ago
JD_PMJD_PM
1846
$endgroup$
add a comment |
$begingroup$
Imagine we deal with a new kind of matter, whose state is described by:
$$PV = AT^3$$
Its internal energy is given by:
$$U = BT^n lnleft(frac{V}{V_0}right) + f(T)$$
Where $A, B$ and $V_0$ is a constant and $f(T)$ is a polynomial function.
Find B and n.
This is what I know:
The given expressions remind me of adiabatic compression/expansion. If we assume quasistatic adiabatic compression/expansion we know that heat won't get out/in the system.
$$Delta U = -W$$
And work is:
$$W = -PDelta V$$
Some thoughts on how to solve the problem
We notice here that we are dealing with a non-ideal gas. Assuming that the above equations are correct and using first thermodynamics law one gets:
$$mathrm{d}U = [nBT^{n-1} lnleft(frac{V}{V_0}right) + f'(T)]mathrm{d}T$$
$$[nBT^{n-1} lnleft(frac{V}{V_0}right) + f'(T)]mathrm{d}T = frac{AT^3}{V}mathrm{d}V$$
I do not see a solution to this differential equation.
There has to be a easier way to get both $B$ and $n$ but how?
Thanks.
thermodynamics equation-of-state
edited 2 hours ago
Gaurang Tandon
5,46262864
asked 3 hours ago
JD_PMJD_PM
1846
$endgroup$
1
$begingroup$
Your differential equation doesn't make sense to me: you have infinitesimals ($mathrm dU$ and $mathrm df$) and finite quantities ($nBT^{n-1}ln(V/V_0)$) being added together. I gather you were trying to differentiate by $T$ throughout?
$endgroup$
– orthocresol♦
2 hours ago
$begingroup$
@orthocresol that is a typo let me fix it.
$endgroup$
– JD_PM
2 hours ago
$begingroup$
Yes I differentiated $U$ with respect to $T$. The idea is to set up a differential equation that relates the change in temperature and volume during the compression/expansion process. I assumed it will be adiabatic (based on the given equation: $PV = AT^3$)
$endgroup$
– JD_PM
2 hours ago
1
$begingroup$
By using the equation $ big( frac{partial U}{partial V}big)_T = Tbig(frac{partial P}{partial T}big)_V - P $, you will get $B$ as $2A$ and $n=3$.
$endgroup$
– Soumik Das
2 hours ago
add a comment |
$begingroup$
Imagine we deal with a new kind of matter, whose state is described by:
$$PV = AT^3$$
Its internal energy is given by:
$$U = BT^n lnleft(frac{V}{V_0}right) + f(T)$$
Where $A, B$ and $V_0$ is a constant and $f(T)$ is a polynomial function.
Find B and n.
This is what I know:
The given expressions remind me of adiabatic compression/expansion. If we assume quasistatic adiabatic compression/expansion we know that heat won't get out/in the system.
$$Delta U = -W$$
And work is:
$$W = -PDelta V$$
Some thoughts on how to solve the problem
We notice here that we are dealing with a non-ideal gas. Assuming that the above equations are correct and using first thermodynamics law one gets:
$$mathrm{d}U = [nBT^{n-1} lnleft(frac{V}{V_0}right) + f'(T)]mathrm{d}T$$
$$[nBT^{n-1} lnleft(frac{V}{V_0}right) + f'(T)]mathrm{d}T = frac{AT^3}{V}mathrm{d}V$$
I do not see a solution to this differential equation.
There has to be a easier way to get both $B$ and $n$ but how?
Thanks.
thermodynamics equation-of-state
edited 2 hours ago
Gaurang Tandon
5,46262864
asked 3 hours ago
JD_PMJD_PM
1846
$endgroup$
Imagine we deal with a new kind of matter, whose state is described by:
$$PV = AT^3$$
Its internal energy is given by:
$$U = BT^n lnleft(frac{V}{V_0}right) + f(T)$$
Where $A, B$ and $V_0$ is a constant and $f(T)$ is a polynomial function.
Find B and n.
This is what I know:
The given expressions remind me of adiabatic compression/expansion. If we assume quasistatic adiabatic compression/expansion we know that heat won't get out/in the system.
$$Delta U = -W$$
And work is:
$$W = -PDelta V$$
Some thoughts on how to solve the problem
We notice here that we are dealing with a non-ideal gas. Assuming that the above equations are correct and using first thermodynamics law one gets:
$$mathrm{d}U = [nBT^{n-1} lnleft(frac{V}{V_0}right) + f'(T)]mathrm{d}T$$
$$[nBT^{n-1} lnleft(frac{V}{V_0}right) + f'(T)]mathrm{d}T = frac{AT^3}{V}mathrm{d}V$$
I do not see a solution to this differential equation.
There has to be a easier way to get both $B$ and $n$ but how?
Thanks.
thermodynamics equation-of-state
thermodynamics equation-of-state
edited 2 hours ago
Gaurang Tandon
5,46262864
asked 3 hours ago
JD_PMJD_PM
1846
edited 2 hours ago
Gaurang Tandon
5,46262864
asked 3 hours ago
JD_PMJD_PM
1846
edited 2 hours ago
Gaurang Tandon
5,46262864
edited 2 hours ago
Gaurang Tandon
5,46262864
edited 2 hours ago
Gaurang Tandon
5,46262864
5,46262864
asked 3 hours ago
JD_PMJD_PM
1846
asked 3 hours ago
JD_PMJD_PM
1846
asked 3 hours ago
JD_PMJD_PM
1846
1846
1
$begingroup$
Your differential equation doesn't make sense to me: you have infinitesimals ($mathrm dU$ and $mathrm df$) and finite quantities ($nBT^{n-1}ln(V/V_0)$) being added together. I gather you were trying to differentiate by $T$ throughout?
$endgroup$
– orthocresol♦
2 hours ago
$begingroup$
@orthocresol that is a typo let me fix it.
$endgroup$
– JD_PM
2 hours ago
$begingroup$
Yes I differentiated $U$ with respect to $T$. The idea is to set up a differential equation that relates the change in temperature and volume during the compression/expansion process. I assumed it will be adiabatic (based on the given equation: $PV = AT^3$)
$endgroup$
– JD_PM
2 hours ago
1
$begingroup$
By using the equation $ big( frac{partial U}{partial V}big)_T = Tbig(frac{partial P}{partial T}big)_V - P $, you will get $B$ as $2A$ and $n=3$.
$endgroup$
– Soumik Das
2 hours ago
add a comment |
1
$begingroup$
Your differential equation doesn't make sense to me: you have infinitesimals ($mathrm dU$ and $mathrm df$) and finite quantities ($nBT^{n-1}ln(V/V_0)$) being added together. I gather you were trying to differentiate by $T$ throughout?
$endgroup$
– orthocresol♦
2 hours ago
$begingroup$
@orthocresol that is a typo let me fix it.
$endgroup$
– JD_PM
2 hours ago
$begingroup$
Yes I differentiated $U$ with respect to $T$. The idea is to set up a differential equation that relates the change in temperature and volume during the compression/expansion process. I assumed it will be adiabatic (based on the given equation: $PV = AT^3$)
$endgroup$
– JD_PM
2 hours ago
1
$begingroup$
By using the equation $ big( frac{partial U}{partial V}big)_T = Tbig(frac{partial P}{partial T}big)_V - P $, you will get $B$ as $2A$ and $n=3$.
$endgroup$
– Soumik Das
2 hours ago
1
1
$begingroup$
Your differential equation doesn't make sense to me: you have infinitesimals ($mathrm dU$ and $mathrm df$) and finite quantities ($nBT^{n-1}ln(V/V_0)$) being added together. I gather you were trying to differentiate by $T$ throughout?
$endgroup$
– orthocresol♦
2 hours ago
$begingroup$
Your differential equation doesn't make sense to me: you have infinitesimals ($mathrm dU$ and $mathrm df$) and finite quantities ($nBT^{n-1}ln(V/V_0)$) being added together. I gather you were trying to differentiate by $T$ throughout?
$endgroup$
– orthocresol♦
2 hours ago
$begingroup$
@orthocresol that is a typo let me fix it.
$endgroup$
– JD_PM
2 hours ago
$begingroup$
@orthocresol that is a typo let me fix it.
$endgroup$
– JD_PM
2 hours ago
$begingroup$
Yes I differentiated $U$ with respect to $T$. The idea is to set up a differential equation that relates the change in temperature and volume during the compression/expansion process. I assumed it will be adiabatic (based on the given equation: $PV = AT^3$)
$endgroup$
– JD_PM
2 hours ago
$begingroup$
Yes I differentiated $U$ with respect to $T$. The idea is to set up a differential equation that relates the change in temperature and volume during the compression/expansion process. I assumed it will be adiabatic (based on the given equation: $PV = AT^3$)
$endgroup$
– JD_PM
2 hours ago
1
1
$begingroup$
By using the equation $ big( frac{partial U}{partial V}big)_T = Tbig(frac{partial P}{partial T}big)_V - P $, you will get $B$ as $2A$ and $n=3$.
$endgroup$
– Soumik Das
2 hours ago
$begingroup$
By using the equation $ big( frac{partial U}{partial V}big)_T = Tbig(frac{partial P}{partial T}big)_V - P $, you will get $B$ as $2A$ and $n=3$.
$endgroup$
– Soumik Das
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You need to use the equation $$left(frac{partial U}{partial V}right)_T=-left[P-Tleft(frac{partial P}{partial T}right)_Vright]$$
answered 2 hours ago
Chet MillerChet Miller
6,8011713
$endgroup$
$begingroup$
Thanks for the answer Chet Millet. However, I still do not see how to derive it. I know the equation you propose comes from the thermodynamic identity: $dU = TdS - PdV + mu dN$. I have been trying to derive it with respect to $V$ but I do not get rid of the entropy term.
$endgroup$
– JD_PM
1 hour ago
1
$begingroup$
The derivation is in every thermo book. You are correct about the starting point. The next step is to express dS in terms of dT and dV. Then, for the dV term, you derive a Maxwell relationship from dA=-SdT-PdV.
$endgroup$
– Chet Miller
59 mins ago
add a comment |
$begingroup$
I am sincerely SORRY that I couldn't provide a LaTeX markup for the answer , but I am busy preparing for one of the biggest exam of my life , I would surely update the answer as soon as the exam is over , any help with OCring image would be appreciated.
Edit 1 : theirs a typo in the equation of dU it's PdV instead of VdP
answered 36 mins ago
Advil SellAdvil Sell
391113
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You need to use the equation $$left(frac{partial U}{partial V}right)_T=-left[P-Tleft(frac{partial P}{partial T}right)_Vright]$$
answered 2 hours ago
Chet MillerChet Miller
6,8011713
$endgroup$
$begingroup$
Thanks for the answer Chet Millet. However, I still do not see how to derive it. I know the equation you propose comes from the thermodynamic identity: $dU = TdS - PdV + mu dN$. I have been trying to derive it with respect to $V$ but I do not get rid of the entropy term.
$endgroup$
– JD_PM
1 hour ago
1
$begingroup$
The derivation is in every thermo book. You are correct about the starting point. The next step is to express dS in terms of dT and dV. Then, for the dV term, you derive a Maxwell relationship from dA=-SdT-PdV.
$endgroup$
– Chet Miller
59 mins ago
add a comment |
$begingroup$
You need to use the equation $$left(frac{partial U}{partial V}right)_T=-left[P-Tleft(frac{partial P}{partial T}right)_Vright]$$
answered 2 hours ago
Chet MillerChet Miller
6,8011713
$endgroup$
$begingroup$
Thanks for the answer Chet Millet. However, I still do not see how to derive it. I know the equation you propose comes from the thermodynamic identity: $dU = TdS - PdV + mu dN$. I have been trying to derive it with respect to $V$ but I do not get rid of the entropy term.
$endgroup$
– JD_PM
1 hour ago
1
$begingroup$
The derivation is in every thermo book. You are correct about the starting point. The next step is to express dS in terms of dT and dV. Then, for the dV term, you derive a Maxwell relationship from dA=-SdT-PdV.
$endgroup$
– Chet Miller
59 mins ago
add a comment |
$begingroup$
You need to use the equation $$left(frac{partial U}{partial V}right)_T=-left[P-Tleft(frac{partial P}{partial T}right)_Vright]$$
answered 2 hours ago
Chet MillerChet Miller
6,8011713
$endgroup$
You need to use the equation $$left(frac{partial U}{partial V}right)_T=-left[P-Tleft(frac{partial P}{partial T}right)_Vright]$$
answered 2 hours ago
Chet MillerChet Miller
6,8011713
answered 2 hours ago
Chet MillerChet Miller
6,8011713
answered 2 hours ago
Chet MillerChet Miller
6,8011713
answered 2 hours ago
Chet MillerChet Miller
6,8011713
6,8011713
$begingroup$
Thanks for the answer Chet Millet. However, I still do not see how to derive it. I know the equation you propose comes from the thermodynamic identity: $dU = TdS - PdV + mu dN$. I have been trying to derive it with respect to $V$ but I do not get rid of the entropy term.
$endgroup$
– JD_PM
1 hour ago
1
$begingroup$
The derivation is in every thermo book. You are correct about the starting point. The next step is to express dS in terms of dT and dV. Then, for the dV term, you derive a Maxwell relationship from dA=-SdT-PdV.
$endgroup$
– Chet Miller
59 mins ago
add a comment |
$begingroup$
Thanks for the answer Chet Millet. However, I still do not see how to derive it. I know the equation you propose comes from the thermodynamic identity: $dU = TdS - PdV + mu dN$. I have been trying to derive it with respect to $V$ but I do not get rid of the entropy term.
$endgroup$
– JD_PM
1 hour ago
1
$begingroup$
The derivation is in every thermo book. You are correct about the starting point. The next step is to express dS in terms of dT and dV. Then, for the dV term, you derive a Maxwell relationship from dA=-SdT-PdV.
$endgroup$
– Chet Miller
59 mins ago
$begingroup$
Thanks for the answer Chet Millet. However, I still do not see how to derive it. I know the equation you propose comes from the thermodynamic identity: $dU = TdS - PdV + mu dN$. I have been trying to derive it with respect to $V$ but I do not get rid of the entropy term.
$endgroup$
– JD_PM
1 hour ago
$begingroup$
Thanks for the answer Chet Millet. However, I still do not see how to derive it. I know the equation you propose comes from the thermodynamic identity: $dU = TdS - PdV + mu dN$. I have been trying to derive it with respect to $V$ but I do not get rid of the entropy term.
$endgroup$
– JD_PM
1 hour ago
1
1
$begingroup$
The derivation is in every thermo book. You are correct about the starting point. The next step is to express dS in terms of dT and dV. Then, for the dV term, you derive a Maxwell relationship from dA=-SdT-PdV.
$endgroup$
– Chet Miller
59 mins ago
$begingroup$
The derivation is in every thermo book. You are correct about the starting point. The next step is to express dS in terms of dT and dV. Then, for the dV term, you derive a Maxwell relationship from dA=-SdT-PdV.
$endgroup$
– Chet Miller
59 mins ago
add a comment |
$begingroup$
I am sincerely SORRY that I couldn't provide a LaTeX markup for the answer , but I am busy preparing for one of the biggest exam of my life , I would surely update the answer as soon as the exam is over , any help with OCring image would be appreciated.
Edit 1 : theirs a typo in the equation of dU it's PdV instead of VdP
answered 36 mins ago
Advil SellAdvil Sell
391113
$endgroup$
add a comment |
$begingroup$
I am sincerely SORRY that I couldn't provide a LaTeX markup for the answer , but I am busy preparing for one of the biggest exam of my life , I would surely update the answer as soon as the exam is over , any help with OCring image would be appreciated.
Edit 1 : theirs a typo in the equation of dU it's PdV instead of VdP
answered 36 mins ago
Advil SellAdvil Sell
391113
$endgroup$
add a comment |
$begingroup$
I am sincerely SORRY that I couldn't provide a LaTeX markup for the answer , but I am busy preparing for one of the biggest exam of my life , I would surely update the answer as soon as the exam is over , any help with OCring image would be appreciated.
Edit 1 : theirs a typo in the equation of dU it's PdV instead of VdP
answered 36 mins ago
Advil SellAdvil Sell
391113
$endgroup$
I am sincerely SORRY that I couldn't provide a LaTeX markup for the answer , but I am busy preparing for one of the biggest exam of my life , I would surely update the answer as soon as the exam is over , any help with OCring image would be appreciated.
Edit 1 : theirs a typo in the equation of dU it's PdV instead of VdP
answered 36 mins ago
Advil SellAdvil Sell
391113
answered 36 mins ago
Advil SellAdvil Sell
391113
answered 36 mins ago
Advil SellAdvil Sell
391113
answered 36 mins ago
Advil SellAdvil Sell
391113
391113
add a comment |
add a comment |
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1
$begingroup$
Your differential equation doesn't make sense to me: you have infinitesimals ($mathrm dU$ and $mathrm df$) and finite quantities ($nBT^{n-1}ln(V/V_0)$) being added together. I gather you were trying to differentiate by $T$ throughout?
$endgroup$
– orthocresol♦
2 hours ago
$begingroup$
@orthocresol that is a typo let me fix it.
$endgroup$
– JD_PM
2 hours ago
$begingroup$
Yes I differentiated $U$ with respect to $T$. The idea is to set up a differential equation that relates the change in temperature and volume during the compression/expansion process. I assumed it will be adiabatic (based on the given equation: $PV = AT^3$)
$endgroup$
– JD_PM
2 hours ago
1
$begingroup$
By using the equation $ big( frac{partial U}{partial V}big)_T = Tbig(frac{partial P}{partial T}big)_V - P $, you will get $B$ as $2A$ and $n=3$.
$endgroup$
– Soumik Das
2 hours ago