A thin partition divides a thermally insulated vessel into a lower compartment of volume V and an upper compartment of volume 2V. The lower compartment contains n moles of an ideal gas; the upper part is evacuated.When the partition is removed, the gas expands and fills both compartments. How many moles n of gas were initially contained in the lower compartment if the entropy change of the gas in this free-expansion process is 17.28 J/K?

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olasunkanmilesanmi olasunkanmilesanmi · Answer 1 · 2019-12-10T23:30:01+01:00

Answer:

1.89mol

Explanation:

The entropy change during free expansion is express as

[tex]S_{f}-S_{i}=nRln(\frac{V_{F}}{V_{I}})\\[/tex]

Where S is the entropy of the system,

n is the amount of mole

R is the gas constant = 8.314 and

V is the volume occupied at the initial and final stage

since the process is n adiabatic free expansion, the entropy of the system is constant. Hence we can re-write the equation as

[tex]S=nRln(\frac{V_{F}}{V_{I}})\\[/tex]

where the [tex]V_{i}=v\\[/tex] and [tex]V_{f}=2v+v=3v\\[/tex]

[tex]S=17.28J/k\\[/tex] and

[tex]R=8.314\\[/tex]

Now if we substitute in values we arrive at

[tex]17.28=(8.314)n*ln(\frac{3v}{v} )\\17.28=(8.314)n*ln(3 )\\17.28=(8.314)n*1.0986\\n=\frac{17.28}{8.314*1.0989}\\n=1.89 mole\\[/tex]