1 The change in the Gibbs energy of a certain constant-pressure process was found to fit the&nbsp;expression ∆𝐺 / 𝐽 = −85.40 + 36.5𝑇 where 𝑇 is the temperature in Kelvins. Find the values ofΔ𝐻 (in 𝐽) and Δ𝑆 (in 𝐽 · 𝐾−1) for the process.&nbsp;4. A Carnot cycle uses 1.00 𝑚𝑜𝑙 of a monatomic ideal gas as a working substance. From an&nbsp;

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1 The change in the Gibbs energy of a certain constant-pressure process was found to fit the&amp;nbsp;expression ∆𝐺 / 𝐽 = −85.40 + 36.5𝑇 where 𝑇 is the temperature in Kelvins. Find the values ofΔ𝐻 (in 𝐽) and Δ𝑆 (in 𝐽 · 𝐾−1) for the process.&amp;nbsp;4. A Carnot cycle uses 1.00 𝑚𝑜𝑙 of a monatomic ideal gas as a working substance. From an&amp;nbsp;

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We are given the expression for the change in Gibbs energy of a constant-pressure process as:ΔG/J=−85.40+36.5Twhere T is the temperature in Kelvin.We know that the relationship between Gibbs energy, enthalpy, and entropy is given by:ΔG=ΔH−TΔSWe can rearrange this equation to solve for ΔH:ΔH=ΔG+TΔSSubstituting the expression for ΔG from the given equation, we get:ΔH=(−85.40+36.5T)J+TΔSWe also know that at constant pressure, the relationship between ΔH and ΔS is given by:ΔH=TΔSSubstituting this into the expression for ΔH, we get:ΔH=TΔS=(−85.40+36.5T)J+TΔSSimplifying, we get:(1−T)ΔS=−85.40JSolving for ΔS, we get:ΔS=−85.40J1−TSubstituting this into the equation for ΔH, we get:ΔH=TΔS=T·−85.40J1−T=−85.40JT1−TTherefore, the values of ΔH and ΔS for the process are ΔH=−85.40JT1−T and ΔS=−85.40J1−T.

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