Chemical Thermodynamics: Entropy Change from Volume Change

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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 Entropy Change from Volume Change.

Name	Status	Author	Last Modified
Chemical Thermodynamics: Entropy Change from Temperature Change	Ready to use	Frances Docherty	30/03/2025 17:56
Chemical Thermodynamics: Entropy Change from Volume Change	Ready to use	Frances Docherty	30/03/2025 18:00

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Question statement

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			Name	Type	Generated Value

mass	number	78.2
gfm	number	28.02
mol	number	2.7908636688
entro	number	16.0832607619

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What is the definition of heat capacity?

Gap 1.

Calculate the change in entropy of the system when $\mathrm{\var{mass}\space g}$ of nitrogen gas at $\mathrm{298 \space K}$ and $\mathrm{1 \space bar}$ pressure doubles its volume in an isothermal, reversible expansion.

Gap 0. $\mathrm{J\space K^{-1}}$ .

Advice

The entropy change resulting from a change in volume of a gas is given by the equation:

$\mathrm{\Delta S=nRln(\frac{V_2}{V_1})}$ ,

where $\mathrm{n}$ is the number of moles of the gas; $\mathrm{R=8.314 J \space K^{-1} \space mol^{-1}}$ is the universal gas constant; $\mathrm{V_1}$ is the initial volume of the gas; and $\mathrm{V_2}$ is the final volume of the gas. The units of the two volumes do not matter, as long as they are the same unit, e.g. both $\mathrm{L}$ , both $\mathrm{mL}$ , both $\mathrm{m^3}$ , etc.

We are told that the volume of our nitrogen gas is doubled. This means that we can express our final volume, $\mathrm{V_2}$ , in terms of our initial volume, $\mathrm{V_1}$ , as:

$\mathrm{V_2=2V_1}$ .

This will prove very useful in our equation, where we will simplify the fraction $\mathrm{\frac{V_2}{V_1}}$ as such:

$\mathrm{\frac{V_2}{V_1}=\frac{2V_1}{V_1}=2}$ .

We are also given the mass of the nitrogen gas, $\mathrm{m=\var{mass}\space g}$ - to use our equation, we must first obtain the number of moles of nitrogen gas, which we can calculate from the mass:

$\mathrm{n=\frac{m}{gFM}=\frac{\var{mass}}{\var{gFM}}=(\var{sigformat(mol,5)}\cdots)\space mol}$ .

Note: We have used the gram formula mass for $\mathrm{N_2}$ , $\mathrm{gFM=28.02\space g}$ , not that of a single $\mathrm{N}$ atom, which would be half of that value. This is because nitrogen gas exists naturally as diatomic $\mathrm{N_2}$ molecules; you will likely remember that during the boiling of liquid nitrogen, the intermolecular attractions between $\mathrm{N_2}$ molecules are overcome but the covalent bonds within the molecules remain intact.

We can now subsitute our values into our original equation (after a applying our volume trick) to obtain the entropy change:

$\mathrm{\Delta S=nRln(\frac{V_2}{V_1})=nRln(\frac{2V_1}{V_1})=nRln(2)}$

$\mathrm{\Rightarrow\space\space\space \Delta S=(\var{sigformat(mol,5)}\cdots)\times 8.314\times ln(2)=\var{sigformat(entro,3)}\space J \space K^{-1}}$ .

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Chemical Thermodynamics: Entropy Change from Volume Change

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Frances Docherty 4 weeks, 1 day ago

Frances Docherty 4 weeks, 1 day ago

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Chemical Thermodynamics: Entropy Change from Volume Change

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`