Chemical Thermodynamics: Gibbs Free Energy Change from Pressure Change 1

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Frances Docherty

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Frances Docherty 4 weeks, 1 day ago

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Created this as a copy of Gibbs Free Energy Change from Pressure Change - Alt (O).

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	2.6
pchange	number	0.9
p2	number	3.5
temp	integer	444
gibbschange	number	1097.2790297539
gckj	number	1.0972790298

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Generated value: `number`

2.6

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gibbschange
p2

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One mole of an unknown gas at $\mathrm{\var{temp}\space K}$ and $\mathrm{\var{p1} \space atm}$ is increased in pressure, undergoing a Gibbs free energy change of $\mathrm{+\var{dpformat(gckj,3)}\space kJ\space mol^{-1}}$ .

Assuming the temperature remains constant throughout, what is the final pressure of the gas?

Unnamed gap $\mathrm{atm}$

Advice

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, $\mathrm{p_1=\var{p1}\space atm}$ , and the Gibbs free energy change associated with the process, $\mathrm{\Delta G=+\var{dpformat(gckj,3)}}$ , which we convert by multiplying by one thousand, to $\mathrm{\Delta G^\circ =\var{dpformat(gibbschange,0)} \space J \space mol^{-1}}$ .

To calculate the final pressure of the gas, we must rearrange our original equation, then substitute in our values:

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

$\mathrm{\Rightarrow\space\space\space ln(\frac{p_2}{p_1})=\frac{\Delta G}{RT}}$

$\mathrm{\Rightarrow\space\space\space \frac{p_2}{p_1}=exp(\frac{\Delta G}{RT})}$

$\mathrm{\Rightarrow\space\space\space {p_2}=p_1exp(\frac{\Delta G}{RT})}$

$\mathrm{\Rightarrow\space\space\space {p_2}=\var{p1}\times exp(\frac{\var{dpformat(gibbschange,3)}}{8.314\times \var{temp}})=\var{sigformat(p2,3)}\space atm}$ .

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Chemical Thermodynamics: Gibbs Free Energy Change from Pressure Change 1

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Frances Docherty 4 weeks, 1 day ago

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Chemical Thermodynamics: Gibbs Free Energy Change from Pressure Change 1

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`