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Standard XII

Chemistry

Ideal Gas Equation

Exactly 100 m...

Question

Exactly 100 mL of oxygen is collected over water of $13^{o} C$ and 800 mm pressure. Calculate the volume of dry oxygen at NTP. (Vapour pressure of water at $23^{o} C$ is 21 mm)

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Solution

$P_{g a s} = P_{a t m} - P_{w a t e r} = 800 - 21 P_{g a s} = 779 m m .$
Now combination of boyles law & charls law.
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}} V_{2} = \frac{100 \times 779 \times 293}{296 \times 760} V_{2} = 101.46 m m .$
volume of dry oxygen at NTP.

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Exactly 100 mL of oxygen is collected over water of 13oC and 800 mm pressure. Calculate the volume of dry oxygen at NTP. (Vapour pressure of water at 23oC is 21 mm)

Pgas=Patm−Pwater=800−21Pgas=779mm. Now combination of boyles law & charls law.P1V1T1=P2V2T2V2=P1V1T2T1P2V2=100×779×293296×760V2=101.46mm. volume of dry oxygen at NTP.

Exactly 100 mL of oxygen is collected over water of $13^{o} C$ and 800 mm pressure. Calculate the volume of dry oxygen at NTP. (Vapour pressure of water at $23^{o} C$ is 21 mm)