A container of $1.0 L$ capacity filled with $1.0$ mole of ideal gas is connected to an evacuated vessel of $9.0 L$ . Calculate change in entropy : $(R = 1.987 c a l .)$

0.188 c a l K^{- 1} {m o l}^{- 1}

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0.4576 c a l K^{- 1} {m o l}^{- 1}

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4.576 c a l K^{- 1} {m o l}^{- 1}

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4.366 c a l K^{- 1} {m o l}^{- 1}

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Solution

The correct option is C $4.576 c a l K^{- 1} {m o l}^{- 1}$
Change in entropy $= n R ln \frac{V_{2}}{V_{1}}$
$= 2.303 \times 1 \times 1.987 log \frac{10}{1}$
$= 2.303 \times 1 \times 1.987$
$= 2.303 \times 1.987$
Change in entropy $= 4.576 c a l K^{- 1} {m o l}^{- 1}$

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A container of 1.0L capacity filled with 1.0 mole of ideal gas is connected to an evacuated vessel of 9.0L. Calculate change in entropy :(R=1.987cal.)

The correct option is C 4.576calK−1mol−1Change in entropy =nRlnV2V1 =2.303×1×1.987log101 =2.303×1×1.987 =2.303×1.987Change in entropy =4.576calK−1mol−1

A container of $1.0 L$ capacity filled with $1.0$ mole of ideal gas is connected to an evacuated vessel of $9.0 L$ . Calculate change in entropy : $(R = 1.987 c a l .)$

The correct option is C $4.576 c a l K^{- 1} {m o l}^{- 1}$
Change in entropy $= n R ln \frac{V_{2}}{V_{1}}$
$= 2.303 \times 1 \times 1.987 log \frac{10}{1}$
$= 2.303 \times 1 \times 1.987$
$= 2.303 \times 1.987$
Change in entropy $= 4.576 c a l K^{- 1} {m o l}^{- 1}$