Calculate the amount of work done by 2 mole of an ideal gas at 298k in reversible isothermal expansion from 10 liter to 20 liter.

-3435 J

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3435 J

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-3400 J

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3400 J

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Solution

The correct option is A
-3435 J

Amount of work done in reversible isothermal expansion can be given as:

$w = - 2.303 n R T l o g (\frac{V_{2}}{V_{1}})$

$∴ w = - 2.303 \times 2 \times 8.314 \times 298 l o g (\frac{20}{10})$

$∴ w = - 3435 J$ work done by the system

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Similar questions

The amount work done by 2 moles of an ideal gas at 298 K in reversible isothermal expansion from 10 litres to 20 litres is.

2

298

10

20

The amount of work done by $2$ mole of an ideal gas at $298$ K in reversible isothermal expansion from $10$ litre to $20$ litre is:

2

298 K

10

20

1 m o l e

10 l

20 l

25^{o} C

(R = 8 J K^{- 1} m o l^{- 1})

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