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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

5.00 m o l

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4.72 m o l

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3.33 m o l

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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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Similar questions

15 m^{3} t o 30 m^{3}

300 K

8.595 k J

J K^{- 1} m o l^{- 1}

10 m^{3} t o 20 m^{3}

300 K

4.754 k J

5 m^{3}

10 m^{3}

(O_{2})

250 J m o l^{- 1} K^{- 1} .

△ G

R = 8.314 J m o l^{- 1} K^{- 1}, l o g e = 2.303)

10

10 atm

1 atm

300 K

R = 2 {cal K}^{- 1} {mol}^{- 1}

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