A certain quantity of a gas occupies 100 mL when collected over water at 15-C and 750 mm pressure- It occupied 91-9 mL in dry state at NTP- Find the aqueous vapour pressure at 15-C-

Let the aqueous vapour pressure be p mm . #160;Initial conditions : P_1 (dry gas) = (750 p)mm #160; V_1 = 100 mL #160; T_1 = 15 ...

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Standard X

Chemistry

Gay Lussac's Law

A certain qua...

Question

A certain quantity of a gas occupies $100 m L$ when collected over water at $15^{\circ} C$ and $750 m m$ pressure. It occupied $91.9 m L$ in dry state at NTP. Find the aqueous vapour pressure at $15^{\circ} C$ .

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Solution

Let the aqueous vapour pressure be $p m m$ .

Initial conditions :

$P_{1}$ (dry gas) $= (750 - p) m m$

$V_{1} = 100 m L$
$T_{1} = 15 + 273 = 288 K$

NTP conditions :

$P_{2} = 760 m m$

$V_{2} = 91.9 m L$

$T_{2} = 293 K$

Applying gas equation we get,

$\frac{(750 - p) \times 100}{288} = \frac{760 \times 91.9}{293}$

$750 - p = \frac{760 \times 91.9 \times 288}{100 \times 293} = 686.5 m m$

$p = 750 - 686.5 = 63.5 m m$ .

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A certain quantity of a gas occupies 100 mL when collected over water at 15∘C and 750 mm pressure. It occupied 91.9 mL in dry state at NTP. Find the aqueous vapour pressure at 15∘C.

Let the aqueous vapour pressure be p mm. Initial conditions :P1 (dry gas) =(750−p)mm V1=100 mLT1=15+273=288 K NTP conditions :P2=760 mm V2=91.9 mL T2=293 K Applying gas equation we get,(750−p)×100288=760×91.9293750−p=760×91.9×288100×293=686.5 mmp=750−686.5=63.5 mm.

A certain quantity of a gas occupies $100 m L$ when collected over water at $15^{\circ} C$ and $750 m m$ pressure. It occupied $91.9 m L$ in dry state at NTP. Find the aqueous vapour pressure at $15^{\circ} C$ .