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Work Done in Isothermal Reversible Process

What is the w...

Question

What is the work which could be obtained from an isothermal reversible expansion of $1 m o l$ of $C l_{2}$ from 1 $d m^{3}$ to 50 $d m^{3}$ at $273 K$ using ideal gas behaviour
(Given $l o g 5 = 0.693$ )

8.88 k J m o l^{- 1}

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- 8.88 k J m o l^{- 1}

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- 88.8 k J m o l^{- 1}

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- 88.8 k J m o l^{- 1}

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Solution

The correct option is B $- 8.88 k J m o l^{- 1}$
Work done for an isothermal reversible process is given as:
$w = - n R T ln \frac{V_{2}}{V_{1}}$
$R = - 8.314 J K^{- 1} m o l^{- 1}$
$T = 273 K$
$V_{1} =$ initial volume $= 1 d m^{3} = 1 L$
$V_{2} =$ final volume $= 50 d m^{3} = 50 L$
therefore
$w = - 8.314 \times 273 \times 2.303 \times (log \frac{50}{1})$
$w = - 8880.80 J m o l^{- 1}$
$w = - 8.88 k J m o l^{- 1}$

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