A certain quantity of a gas occupied 200 mL when collected over water at 15-C and 745 mm pressure- It occupies 182-6 mL in the dry state at NTP- Find the vapour pressure of water at 15-C-

Given, P_moist=745mm Hg P_2=760mm Hg V_1=200ml T_1=288.15K #160; #160; #160; #160; V_2=182.6ml T_2=273.15K PV=nR ...

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Standard XII

Chemistry

Raoult's Law

A certain qua...

Question

A certain quantity of a gas occupied $200 m L$ when collected over water at $15^{\circ} C$ and $745 m m$ pressure. It occupies $182.6 m L$ in the dry state at NTP. Find the vapour pressure of water at $15^{\circ} C$ .

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Solution

Given,
$P_{m o i s t} = 745 m m H g P_{2} = 760 m m H g V_{1} = 200 m l T_{1} = 288.15 K$ $V_{2} = 182.6 m l T_{2} = 273.15 K$
$P V = n R T \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \frac{P \times 200}{288.15} = \frac{760 \times 182.6}{273} P = 732.3865 P_{d r y} = (732.39 \sim 745) = 12.6135 H g$ (ideal gas equation)
Vapor pressure of water at $15^{\circ} C$ is $12.6135 m m H g$

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A certain quantity of a gas occupied 200 mL when collected over water at 15∘C and 745 mm pressure. It occupies 182.6 mL in the dry state at NTP. Find the vapour pressure of water at 15∘C.

Given,Pmoist=745mmHgP2=760mmHgV1=200mlT1=288.15K V2=182.6mlT2=273.15KPV=nRTP1V1T1=P2V2T2P×200288.15=760×182.6273P=732.3865Pdry=(732.39∼745)=12.6135Hg (ideal gas equation)Vapor pressure of water at 15∘C is 12.6135mmHg

A certain quantity of a gas occupied $200 m L$ when collected over water at $15^{\circ} C$ and $745 m m$ pressure. It occupies $182.6 m L$ in the dry state at NTP. Find the vapour pressure of water at $15^{\circ} C$ .