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Byju's Answer

Standard IX

Chemistry

The Ideal Gas Equation

50 cm3 of hyd...

Question

$50 c m^{3}$ of hydrogen is collected over water at $17^{\circ} C$ and 750 mm Hg pressure. Calculate the volume of dry gas at S.T.P. The water vapour pressure at $17^{\circ} C$ is 14 mm Hg.

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Solution

We will use the following relation

$\frac{P 1 V 1}{T 1} = \frac{P 2 V 2}{T 2}$ ------- (1)

We are given that $P_{1}$ = 750 - vapour pressure of the gas

= (750 - 14) mm of Hg

= 736mm of Hg

$T_{1}$ = 170 C = (273+ 17) K = 290K

$V_{1}$ = 50 cm3

We know that at STP conditions pressure = 1atm = 760 mm of Hg and temperature = 273 K

Hence $P_{2}$ = 760 mm of Hg and $T_{2}$ = 273 K

Substituting the above values in equation (1), we get

V2 = $\frac{736 \times 50 \times 273}{760 \times 290}$

= 45.58 $c m^{3}$

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50cm3 of hydrogen is collected over water at 17∘C and 750 mm Hg pressure. Calculate the volume of dry gas at S.T.P. The water vapour pressure at 17∘C is 14 mm Hg.

$50 c m^{3}$ of hydrogen is collected over water at $17^{\circ} C$ and 750 mm Hg pressure. Calculate the volume of dry gas at S.T.P. The water vapour pressure at $17^{\circ} C$ is 14 mm Hg.