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Why does the negative sign arise in this thermodynamic relation?

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Why does the negative sign arise in this thermodynamic relation?

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1

$begingroup$

I can't understand why $left(frac{partial P}{partial V} right)_T=-left(frac{partial P}{partial T} right)_Vleft(frac{partial T}{partial V}right)_P$. Why does the negative sign arise? I can easily write $left(frac{partial P}{partial V}right)_T=left(frac{partial P}{partial T}right)_Vleft(frac{partial T}{partial V}right)_P$ from the rule of partial derivative, but what's the negative sign for?

thermodynamics differentiation

share|cite|improve this question

edited 4 hours ago

ACuriousMind♦

72.5k18126318

asked 4 hours ago

Srijan GhoshSrijan Ghosh

83

New contributor

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$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your second formula is easily written but it is wrong. The first formula is right and to find out why try en.wikipedia.org/wiki/Triple_product_rule
  $endgroup$
  – Andrew Steane
  4 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I can't understand why $left(frac{partial P}{partial V} right)_T=-left(frac{partial P}{partial T} right)_Vleft(frac{partial T}{partial V}right)_P$. Why does the negative sign arise? I can easily write $left(frac{partial P}{partial V}right)_T=left(frac{partial P}{partial T}right)_Vleft(frac{partial T}{partial V}right)_P$ from the rule of partial derivative, but what's the negative sign for?

thermodynamics differentiation

share|cite|improve this question

edited 4 hours ago

ACuriousMind♦

72.5k18126318

asked 4 hours ago

Srijan GhoshSrijan Ghosh

83

New contributor

Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your second formula is easily written but it is wrong. The first formula is right and to find out why try en.wikipedia.org/wiki/Triple_product_rule
  $endgroup$
  – Andrew Steane
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I can't understand why $left(frac{partial P}{partial V} right)_T=-left(frac{partial P}{partial T} right)_Vleft(frac{partial T}{partial V}right)_P$. Why does the negative sign arise? I can easily write $left(frac{partial P}{partial V}right)_T=left(frac{partial P}{partial T}right)_Vleft(frac{partial T}{partial V}right)_P$ from the rule of partial derivative, but what's the negative sign for?

thermodynamics differentiation

share|cite|improve this question

edited 4 hours ago

ACuriousMind♦

72.5k18126318

asked 4 hours ago

Srijan GhoshSrijan Ghosh

83

New contributor

Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 4 hours ago

ACuriousMind♦

72.5k18126318

asked 4 hours ago

Srijan GhoshSrijan Ghosh

83

New contributor

Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 4 hours ago

ACuriousMind♦

72.5k18126318

asked 4 hours ago

Srijan GhoshSrijan Ghosh

83

New contributor

Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 4 hours ago

ACuriousMind♦

72.5k18126318

edited 4 hours ago

ACuriousMind♦

72.5k18126318

asked 4 hours ago

Srijan GhoshSrijan Ghosh

83

New contributor

Srijan Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 4 hours ago

Srijan GhoshSrijan Ghosh

83

2 Answers
2

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2

$begingroup$

This is not a simple application of partial derivatives, since the variables that are being held constant vary here, but an instance of the triple product rule, which says that for any three quantities $x,y,z$ depending on each other, the relation
$$ left(frac{partial x}{partial y}right)_zleft(frac{partial y}{partial z} right)_xleft(frac{partial z}{partial x}right)_y = -1$$
holds.

share|cite|improve this answer

answered 4 hours ago

ACuriousMind♦ACuriousMind

72.5k18126318

$endgroup$

add a comment | 

2

$begingroup$

This all starts from the basic relationship $$dP=left(frac{partial P}{partial T}right)_VdT+left(frac{partial P}{partial V}right)_TdV$$Since, at constant pressure, dP=0, if we solve for dT/dV at constant pressure, we obtain:
$$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$

share|cite|improve this answer

answered 3 hours ago

Chester MillerChester Miller

15.3k2824

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add a comment | 

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2

$begingroup$

This is not a simple application of partial derivatives, since the variables that are being held constant vary here, but an instance of the triple product rule, which says that for any three quantities $x,y,z$ depending on each other, the relation
$$ left(frac{partial x}{partial y}right)_zleft(frac{partial y}{partial z} right)_xleft(frac{partial z}{partial x}right)_y = -1$$
holds.

share|cite|improve this answer

answered 4 hours ago

ACuriousMind♦ACuriousMind

72.5k18126318

$endgroup$

add a comment | 

2

$begingroup$

This is not a simple application of partial derivatives, since the variables that are being held constant vary here, but an instance of the triple product rule, which says that for any three quantities $x,y,z$ depending on each other, the relation
$$ left(frac{partial x}{partial y}right)_zleft(frac{partial y}{partial z} right)_xleft(frac{partial z}{partial x}right)_y = -1$$
holds.

share|cite|improve this answer

answered 4 hours ago

ACuriousMind♦ACuriousMind

72.5k18126318

$endgroup$

add a comment | 

$begingroup$

This is not a simple application of partial derivatives, since the variables that are being held constant vary here, but an instance of the triple product rule, which says that for any three quantities $x,y,z$ depending on each other, the relation
$$ left(frac{partial x}{partial y}right)_zleft(frac{partial y}{partial z} right)_xleft(frac{partial z}{partial x}right)_y = -1$$
holds.

share|cite|improve this answer

answered 4 hours ago

ACuriousMind♦ACuriousMind

72.5k18126318

$endgroup$

share|cite|improve this answer

answered 4 hours ago

ACuriousMind♦ACuriousMind

72.5k18126318

answered 4 hours ago

ACuriousMind♦ACuriousMind

72.5k18126318

answered 4 hours ago

ACuriousMind♦ACuriousMind

72.5k18126318

2

$begingroup$

This all starts from the basic relationship $$dP=left(frac{partial P}{partial T}right)_VdT+left(frac{partial P}{partial V}right)_TdV$$Since, at constant pressure, dP=0, if we solve for dT/dV at constant pressure, we obtain:
$$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$

share|cite|improve this answer

answered 3 hours ago

Chester MillerChester Miller

15.3k2824

$endgroup$

add a comment | 

2

$begingroup$

This all starts from the basic relationship $$dP=left(frac{partial P}{partial T}right)_VdT+left(frac{partial P}{partial V}right)_TdV$$Since, at constant pressure, dP=0, if we solve for dT/dV at constant pressure, we obtain:
$$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$

share|cite|improve this answer

answered 3 hours ago

Chester MillerChester Miller

15.3k2824

$endgroup$

add a comment | 

$begingroup$

This all starts from the basic relationship $$dP=left(frac{partial P}{partial T}right)_VdT+left(frac{partial P}{partial V}right)_TdV$$Since, at constant pressure, dP=0, if we solve for dT/dV at constant pressure, we obtain:
$$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$

share|cite|improve this answer

answered 3 hours ago

Chester MillerChester Miller

15.3k2824

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Chester MillerChester Miller

15.3k2824

answered 3 hours ago

Chester MillerChester Miller

15.3k2824

answered 3 hours ago

Chester MillerChester Miller

15.3k2824