G = H – T S …(i) G = U + P V – T S Δ G = Δ U + P Δ V + V Δ P - T Δ S - S Δ T From the first and second laws, T Δ S = Δ U + P Δ V Δ G = V Δ P - S Δ T At constant pressure, Δ P = 0 Δ G Δ T = – S …(ii) From equations. (i) and (ii) G = H + T Δ G Δ T or G = H + T ∂ G ∂ T P - H T 2 = - G T 2 + 1 T ∂ G ∂ T P = ∂ G / T ∂ T P H = - T 2 ∂ G / T ∂ T P

Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIAdiabatic Process/PolytropicMedium2 minPYQ_2020

ChemistryMediumsingle choice

Enthalpy is equal to

Options:

Answer:

Solution:

$G = H - T S$ …(i)
$G = U + P V - T S$
$Δ G = Δ U + P Δ V + V Δ P - T Δ S - S Δ T$
From the first and second laws,
$T Δ S = Δ U + P Δ V$
$Δ G = V Δ P - S Δ T$
At constant pressure, $Δ P = 0$
$\frac{Δ G}{Δ T} = - S$ …(ii)
From equations. (i) and (ii)
$G = H + T \frac{Δ G}{Δ T}$
or $G = H + T {(\frac{\partial G}{\partial T})}_{P}$
$- \frac{H}{T^{2}} = - \frac{G}{T^{2}} + \frac{1}{T} {(\frac{\partial G}{\partial T})}_{P} = {[\frac{\partial (G / T)}{\partial T}]}_{P}$
$H = - T^{2} {[\frac{\partial (G / T)}{\partial T}]}_{P}$

Stream:NTA_ABHYASSubject:ChemistryTopic:Thermodynamics - IISubtopic:Adiabatic Process/Polytropic

⏱ 2mℹ️ Source: PYQ_2020

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