Chemical Thermodynamics: Gibbs Free Energy Change from Pressure Change 2

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Frances Docherty

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Created this as a copy of Gibbs Free Energy Change from Pressure Change.

Name	Status	Author	Last Modified
Chemical Thermodynamics: Gibbs Free Energy Change from Pressure Change 2	Ready to use	Frances Docherty	30/03/2025 18:17
Chemical Thermodynamics: Gibbs Free Energy Change from Pressure Change 1	Ready to use	Frances Docherty	30/03/2025 18:17

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Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

p1	number	3
pchange	number	0.4
p2	number	3.4
temp	integer	298
gibbschange	number	310.1006984148
gckj	number	0.3101006984

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What is the change in the Gibbs free energy at $\mathrm{\var{temp}\space K}$ when the pressure of a gas is increased from $\mathrm{\var{p1} \space atm}$ by $\mathrm{\var{pchange} \space atm}$ , assuming the temperature remains constant throughout?

Unnamed gap $\mathrm{kJ\space mol^{-1}}$

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The Gibbs free energy change resulting from a change in pressure at constant temperature is given by the equation:

$\mathrm{\Delta G=RTln(\frac{p_2}{p_1})}$ ,

where $\mathrm{R=8.314 J \space K^{-1} \space mol^{-1}}$ is the universal gas constant; $\mathrm{T}$ is the temperature of the gas, in $\mathrm{K}$ ; $\mathrm{p_1}$ is the initial pressure of the gas; and $\mathrm{p_2}$ is the final pressure of the gas. The units of the two pressures do not matter, as long as they are the same unit, e.g. both $\mathrm{Pa}$ , both $\mathrm{torr}$ , both $\mathrm{bar}$ , etc.

We are given the temperature of the system, $\mathrm{T=\var{temp}\space K}$ , as well as the initial gas pressure, $\mathrm{p_1=\var{p1}\space atm}$ . To use our original equation, we must calculate the final gas pressure, which we know is $\mathrm{\var{pchange}\space atm}$ greater than our initial pressure:

$\mathrm{p_2=p_1+\Delta p=\var{p1}+\var{pchange}=\var{p2}\space atm}$ .

We can now subsitute our values into the original equation to obtain the Gibbs free energy change:

$\mathrm{\Delta G=8.314\times\var{temp}\times ln(\frac{\var{p2}}{\var{p1}})=\var{dpformat(gibbschange,1)}\space J \space mol^{-1}}$ .

To quote our answer in the units asked for, we divide by one thousand to obtain:

$\mathrm{\Delta G=\var{sigformat(gckj,3)}\space kJ\space mol^{-1}}$ .

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Chemical Thermodynamics: Gibbs Free Energy Change from Pressure Change 2

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Frances Docherty 4 weeks, 1 day ago

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Frances Docherty 4 weeks, 1 day ago

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Chemical Thermodynamics: Gibbs Free Energy Change from Pressure Change 2

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Frances Docherty 4 weeks, 1 day ago

Frances Docherty 4 weeks, 1 day ago

Frances Docherty 4 weeks, 1 day ago

Error in variable testing condition

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Generated value: number

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Generated value: `number`