Mrthfx

Vocabulary for giving just numbers, not a full answer

Conservation of Mass and Energy

Declaring and defining template, and specialising them

Can one live in the U.S. and not use a credit card?

They call me Inspector Morse

What are some noteworthy "mic-drop" moments in math?

Should I take out a loan for a friend to invest on my behalf?

Do I really need to have a scientific explanation for my premise?

Signed and unsigned numbers

NASA's RS-25 Engines shut down time

Should I tell my boss the work he did was worthless

Single word request: Harming the benefactor

Word for a person who has no opinion about whether god exists

Accepted offer letter, position changed

Why would one plane in this picture not have gear down yet?

Why does Captain Marvel assume the people on this planet know this?

Can you reject a postdoc offer after the PI has paid a large sum for flights/accommodation for your visit?

Reversed Sudoku

Do items de-spawn in Diablo?

PTIJ: Should I kill my computer after installing software?

How can I get players to stop ignoring or overlooking the plot hooks I'm giving them?

How to detect if C code (which needs 'extern C') is compiled in C++

Shifting between bemols (flats) and diesis (sharps)in the key signature

How to draw cubes in a 3 dimensional plane

Why does the negative sign arise in this thermodynamic relation?

Why is the derivative of the Fermi-Dirac distribution negative?Partial derivatives vs total derivatives in thermodynamicsWhy does the material derivative and transport theorem look different?Partial Derivatives Relation on Thermodynamics and minus signExplain me the sign in imageFourier Heat Conduction Law - why is temperature gradient negative?What is the significance of the temperature derivative of the isentropic bulk modulus?Entropy and heat capacity at constant volumeWhy does the internal energy's total differential behave like this?How to know the dependances of thermodynamic variables

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 | 

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

active

oldest

votes

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

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

Srijan Ghosh is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f465965%2fwhy-does-the-negative-sign-arise-in-this-thermodynamic-relation%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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