The correct property relation for enthalpy change dh is:

T d s - v d P

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T d s - P d v

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T d s + v d P

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T d s + P d v

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Solution

The correct option is C $T d s + v d P$
$From Ist law of Thermodynamics$
$Q = d h + W$
$T d s = d h - v d P for open system$
$d h = T d s + v d P$

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Similar questions

Q. The enthalpy change for the process

C (g r a p h i t e) \to C (g)

is called

Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A $\to$ B along one route is $Δ_{r} H$ and $Δ_{r} H a n d Δ_{r} H_{1}, Δ_{r} H_{2}, Δ_{r} H_{3} \dots$ represent enthalpies of intermediate reactions leading to product B. What will be the relation between $Δ_{r} H$ overall reaction and $Δ_{r} H_{1}, Δ_{r} H_{2} \dots$ etc for intermediate reactions.

Q. Match the thermodynamic properties (List I) with their relation (List II).

List I	List II
A. Free energy change ( $Δ G^{o}$ )	(i) $R T {log}_{e} K$
B. Entropy change $Δ S^{o}$	(ii) $- n F E$
C. $Δ H^{o}$ enthalpy change of a reaction in standard state	(iii) ${R T}^{2} {(\frac{d ln K}{d T})}_{P}$
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Q. How many of the following relation(s) is/are correct regarding the given properties?

(a)

C < N > O < F

: First ionization enthalpy

(b)

C l > F > B r

: First electron affinity

(c)

O > S > S e

: Electronegativity

(d)

L i > B e > B

: Second ionization enthalpy

(e)

B > B e > L i

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Q. Which one of the following statements is correct in relation to ionization enthalpy?

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The correct property relation for enthalpy change dh is:

The correct option is C Tds+vdPFrom Ist law of Thermodynamics Q=dh+W Tds=dh−vdP for open system dh=Tds+vdP

The correct option is C $T d s + v d P$
$From Ist law of Thermodynamics$
$Q = d h + W$
$T d s = d h - v d P for open system$
$d h = T d s + v d P$