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Thermodynamic Processes

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Question

What is the work which could be obtained from an isothermal reversible expansion of 1 mol of $C l_{2}$ from $1 d m^{3}$ to $50 d m^{3}$ at 273 K using ideal gas behaviour.

- 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 $8.88 k J m o l^{- 1}$
For an ideal gas
$w = - R T l n \frac{V_{2}}{V_{1}}$
$= - (8.314 J K^{- 1} m o l^{- 1}) (273 K) \times 2.303 \times l o g (\frac{50 d m^{3}}{1 d m^{3}})$
$= - 8882 J m o l^{- 1} = - 8.88 k J m o l^{- 1}$
Work obtained = $8.88 k J m o l^{- 1}$

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Similar questions

Q. What is the work which could be obtained from an isothermal reversible expansion of

1 m o l

C l_{2}

from 1

d m^{3}

to 50

d m^{3}

273 K

using ideal gas behaviour
(Given

l o g 5 = 0.693

)

Q. What is the work which could be obtained from an isothermal reversible expansion of

1 m o l

C l_{2}

from 1

d m^{3}

to 50

d m^{3}

273 K

using ideal gas behaviour
(Given

l o g 5 = 0.693

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Q. What is the work which could be obtained from an isothermal reversible expansion of 1 mol of

C l_{2}

from

1 d m^{3}

50 d m^{3}

at 273 K using ideal gas behaviour.

Calculate the work done for a reversible isothermal expansion of one mole of an ideal gas at $127^{\circ} C$ from a volume of $10 d m^{3}$ to $20 d m^{3}$

Q. For the reversible isothermal expansion of one mole of an ideal gas at

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Thermodynamic Processes

Standard XIII Chemistry

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What is the work which could be obtained from an isothermal reversible expansion of 1 mol of Cl2 from 1 dm3 to 50 dm3 at 273 K using ideal gas behaviour.

The correct option is B 8.88 kJ mol−1For an ideal gas w=−RT lnV2V1 =−(8.314JK−1mol−1)(273K)×2.303×log(50dm31dm3) =−8882 J mol−1=−8.88 kJ mol−1 Work obtained = 8.88 kJ mol−1

What is the work which could be obtained from an isothermal reversible expansion of 1 mol of $C l_{2}$ from $1 d m^{3}$ to $50 d m^{3}$ at 273 K using ideal gas behaviour.

The correct option is B $8.88 k J m o l^{- 1}$
For an ideal gas
$w = - R T l n \frac{V_{2}}{V_{1}}$
$= - (8.314 J K^{- 1} m o l^{- 1}) (273 K) \times 2.303 \times l o g (\frac{50 d m^{3}}{1 d m^{3}})$
$= - 8882 J m o l^{- 1} = - 8.88 k J m o l^{- 1}$
Work obtained = $8.88 k J m o l^{- 1}$