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Charles' Law

Ten moles of ...

Question

Ten moles of an ideal gas at constant temperature 600K is compressed from 100 L to 10 L. The work done in the process is

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Solution

Given,

Initial volume, $V_{1} = 100$

Final Volume, $V_{2} = 10$

Work done, $W = 2.3026 n R T l o g_{10} (\frac{V_{2}}{V_{1}})$

$= 2.3026 \times 10 \times 8.3 \times 600 l o g_{10} (\frac{10}{100}) = 11.4 \times 10^{4} J$

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