For an ideal gas - T- Ps is equal to where all symbols carry their usual meaning

Using maxwell's relation, we have nbsp; (∂ T∂ P)_s=(∂ v∂ s)_P=(∂ v∂ T)_P(∂ T∂ s)_P C_P=T(∂ s∂ T )_P and nbsp; nbsp; β=1v(∂ s∂ T)_P (∂ T∂ P)_s=TvβC_P ...

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Byju's Answer

Standard XII

Chemistry

Heat of Reaction

For an ideal ...

Question

For an ideal gas ${(\frac{\partial T}{\partial P})}_{s}$ is equal to ( where all symbols carry their usual meaning )

\frac{T v β}{C_{P}}

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\frac{T v}{C_{P} β}

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\frac{C_{P}}{T v β}

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\frac{C_{P} β}{T v}

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Solution

The correct option is A $\frac{T v β}{C_{P}}$
Using maxwell's relation, we have
${(\frac{\partial T}{\partial P})}_{s} = {(\frac{\partial v}{\partial s})}_{P} = {(\frac{\partial v}{\partial T})}_{P} {(\frac{\partial T}{\partial s})}_{P}$

$C_{P} = T {(\frac{\partial s}{\partial T})}_{P}$

and $β = \frac{1}{v} {(\frac{\partial s}{\partial T})}_{P}$

${(\frac{\partial T}{\partial P})}_{s} = \frac{T v β}{C_{P}}$

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For an ideal gas (∂T∂P)s is equal to ( where all symbols carry their usual meaning )

The correct option is A TvβCPUsing maxwell's relation, we have (∂T∂P)s=(∂v∂s)P=(∂v∂T)P(∂T∂s)P CP=T(∂s∂T)P and β=1v(∂s∂T)P (∂T∂P)s=TvβCP

For an ideal gas ${(\frac{\partial T}{\partial P})}_{s}$ is equal to ( where all symbols carry their usual meaning )