 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2024/paper-2/a9/f/solution

For fixed atom number, volume scales as the cube of the touching-sphere radius. Compressing silver to the stated gold lattice size therefore requires

\frac{V _{f}}{V _{i}} = (\frac{1.442}{1.444})^{3} = 0.9958506,

a volume decrease of about

0.415%

. If the bulk modulus is constant,

d p = - B \frac{d V}{V}, Δ p = B ln \frac{V _{i}}{V _{f}} = 3 B ln \frac{1.444}{1.442} .

Using the silver value derived above gives

Δ p = 0.1436 GPa ≃ 1.44 \times 1 0^{8} Pa .

For this small compression,

Δ p ≃ - B Δ V / V

gives almost the same estimate. This is an additional pressure relative to the initial standard-pressure state, using the question's free-electron model modulus rather than an empirical silver modulus.

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