A vessel contains air at a temperature of $15^{0} C$ and 60% R.H. What will be the R.H if it is heated to $20^{0} C$ ? (S.V.P at $15^{0} C$ is 12.67 & at $20^{0} C$ is 17.36mm of Hg respectively)

262%

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26.2%

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44.5%

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46.2%

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Solution

The correct option is C 44.5%
By definition,
$\frac{V . P}{S . V . P} \times 100$ = $\frac{60}{100}$ (both VP and SVP are in room temperature)
$= \frac{(V . P)_{15}}{12.67}$
$(V . P)_{15} = 7.6 m m$ $o f$ $H g$
Now since unsaturated vapour obeys gas equation & mass of water remains constant so
$P$ = $(\frac{n R T}{V})$
$\Rightarrow$ $P α T$
$\frac{(V . P)_{15}}{(V . P)_{2} 0}$
$= \frac{273 + 15}{273 + 20}$
$\Rightarrow$ $(R H)_{20}$ = 7.73 mm of Hg
So $(R . H)_{20} = \frac{(V . P)_{20}}{(S . V . P)_{20}} \times 100$
$= 44.5$ %

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A vessel contains air at a temperature of 150C and 60% R.H. What will be the R.H if it is heated to 200C? (S.V.P at 150C is 12.67 & at 200C is 17.36mm of Hg respectively)

A vessel contains air at a temperature of $15^{0} C$ and 60% R.H. What will be the R.H if it is heated to $20^{0} C$ ? (S.V.P at $15^{0} C$ is 12.67 & at $20^{0} C$ is 17.36mm of Hg respectively)