Calculate work done w for the conversion of 0-5 moles of water at 100 -C to steam at 1 atm pressure- Heat of vaporisation of water at 100 -C is 40670 J mol-1

Volume of 0.5 moles of steam at 1 atm pressure = nRTP = 0.5 × 0.082 × 3731.0 L ≈ 15.3 L Change in volume = Vol. of steam Vol. of water = 15.3 negligible = ...

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Standard XI

Chemistry

Enthalpy

Calculate wor...

Question

Calculate work done $w$ for the conversion of $0.5$ moles of water at $100^{\circ} C$ to steam at $1 a t m$ pressure. Heat of vaporisation of water at $100^{\circ} C$ is $40670 J m o l^{- 1}$

- 20.2 L - a t m

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- 21.2 L - a t m

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- 18.3 L - a t m

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- 15.3 L - a t m

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Solution

The correct option is D $- 15.3 L - a t m$
Volume of 0.5 moles of steam at 1 atm pressure
$= \frac{n R T}{P}$
$= \frac{0.5 \times 0.082 \times 373}{1.0} L$
$\approx 15.3 L$
$Change in volume = Vol. of steam - Vol. of water$
$= 15.3 - negligible$
$= 15.3 L$
So, work done by the system,
$W = - P_{e x t} \times volume change$
$= - 1 \times 15.3 L - a t m$
$= - 15.3 L - a t m$

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Calculate work done w for the conversion of 0.5 moles of water at 100 ∘C to steam at 1 atm pressure. Heat of vaporisation of water at 100 ∘C is 40670 Jmol−1

The correct option is D −15.3 L−atmVolume of 0.5 moles of steam at 1 atm pressure =nRTP =0.5×0.082×3731.0 L ≈15.3 L Change in volume = Vol. of steam - Vol. of water = 15.3 - negligible =15.3 L So, work done by the system, W=−Pext×volume change =−1×15.3 L−atm =−15.3 L−atm

Calculate work done $w$ for the conversion of $0.5$ moles of water at $100^{\circ} C$ to steam at $1 a t m$ pressure. Heat of vaporisation of water at $100^{\circ} C$ is $40670 J m o l^{- 1}$