Why does the negative sign arise in this thermodynamic relation?
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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
New contributor
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add a comment |
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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
New contributor
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2
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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
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– Andrew Steane
7 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
New contributor
$endgroup$
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
thermodynamics differentiation
New contributor
New contributor
edited 7 hours ago
ACuriousMind♦
72.5k18126318
72.5k18126318
New contributor
asked 7 hours ago
Srijan GhoshSrijan Ghosh
133
133
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New contributor
2
$begingroup$
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
7 hours ago
add a comment |
2
$begingroup$
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
7 hours ago
2
2
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
add a comment |
2 Answers
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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.
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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:
$$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$
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2 Answers
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$begingroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
answered 7 hours ago
ACuriousMind♦ACuriousMind
72.5k18126318
72.5k18126318
add a comment |
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:
$$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$
$endgroup$
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:
$$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$
$endgroup$
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:
$$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$
$endgroup$
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:
$$left(frac{partial T}{partial V}right)_P=-frac{left(frac{partial P}{partial V}right)_T}{left(frac{partial P}{partial T}right)_V}$$
answered 7 hours ago
Chester MillerChester Miller
15.3k2824
15.3k2824
add a comment |
add a comment |
Srijan Ghosh is a new contributor. Be nice, and check out our Code of Conduct.
Srijan Ghosh is a new contributor. Be nice, and check out our Code of Conduct.
Srijan Ghosh is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
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
7 hours ago