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Work Done in Isothermal Reversible Process

Carbon monoxi...

Question

Carbon monoxide is allowed to expand isothermally and reversibly from $15 m^{3} t o 30 m^{3}$ at $300 K$ and work obtained is $8.595 k J$ . Calculate the number of moles of carbon monoxide. Take R = 8.3 $J K^{- 1} m o l^{- 1}$ and ln2 = 0.69

A

5.00 m o l

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B

4.72 m o l

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C

3.33 m o l

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D

7.82 m o l

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Solution

The correct option is A $5.00 m o l$
Work done in an isothermal reversible process is given as:
$w = - n R T l n \frac{V_{2}}{V_{1}}$
Given: $V_{1} = 15 m^{3}$
$V_{2} = 30 m^{3}$
$T = 300 K$
$R = 8.3 J K^{- 1} m o l^{- 1}$
$W = - 8.585 \times 10^{3} J$
(as work is done by the system)
So,
$- 8.595 \times 10^{3} = - n \times 8.3 \times 300 \times l n \frac{20}{10}$

On solving we get,
$n = 5 m o l$

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