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Standard XII

Chemistry

Heavy Water

One mole of l...

Question

One mole of liquid water at its boiling point vapourises against a constant external pressure of $1 a t m .$ at the same temperature. Assuming ideal behaviour and initial volume of water vapours are zero , the work done by the system nearly :

- 3102 J

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+ 3102 J

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- 4268 J

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+ 4268 J

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Solution

The correct option is B $+ 3102 J$
$V = \frac{n R T}{P}$
$= \frac{1 m o l \times (0.0821 L a t m / m o l k) \times 373 k}{1 a t m}$
$= 30.62 L$
Work done by the system $= P (V_{f} - V_{i})$
$= 1 a t m (30.62 - 0)$
$= 30.62 L a t m$
$= 30.62 \times 101.32 J$
$= + 3102 J$

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One mole of liquid water at its boiling point vapourises against a constant external pressure of 1atm. at the same temperature. Assuming ideal behaviour and initial volume of water vapours are zero , the work done by the system nearly :

The correct option is B +3102JV=nRTP=1mol×(0.0821Latm/molk)×373k1atm=30.62LWork done by the system=P(Vf−Vi)=1atm(30.62−0)=30.62Latm=30.62×101.32J=+3102J

One mole of liquid water at its boiling point vapourises against a constant external pressure of $1 a t m .$ at the same temperature. Assuming ideal behaviour and initial volume of water vapours are zero , the work done by the system nearly :

The correct option is B $+ 3102 J$
$V = \frac{n R T}{P}$
$= \frac{1 m o l \times (0.0821 L a t m / m o l k) \times 373 k}{1 a t m}$
$= 30.62 L$
Work done by the system $= P (V_{f} - V_{i})$
$= 1 a t m (30.62 - 0)$
$= 30.62 L a t m$
$= 30.62 \times 101.32 J$
$= + 3102 J$