Question

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?

Answer 1

Answer:

1.89mol

Explanation:

The entropy change during free expansion is express as

$S_{f}-S_{i}=nRln(\frac{V_{F}}{V_{I}})$

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

$S=nRln(\frac{V_{F}}{V_{I}})$

where the  $V_{i}=v$ and $V_{f}=2v+v=3v$

$S=17.28J/k$ and

$R=8.314$

Now if we substitute in values we arrive at

$17.28=(8.314)n*ln(\frac{3v}{v} )\\17.28=(8.314)n*ln(3 )\\17.28=(8.314)n*1.0986\\n=\frac{17.28}{8.314*1.0989}\\n=1.89 mole$