$1.22 g$ of a gas measured over water at $15^{\circ} C$ and a pressure of $775 m m$ of mercury occupied $900 m L$ . Calculate the volume of dry gas at NTP. Vapor pressure of water at $15^{\circ} C$ is $14 m m$ .

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Solution

Pressure of dry gas $=$ Pressure of moist gas $-$ Aqueous tension
$= 775 - 14$
$= 761 m m$
Initial conditions NTP conditions
$V_{1} = 900 m L$ $V_{2} = ?$
$P_{1} = 761 m m$ $P_{2} = 760 m m$
$T_{1} = (273 + 15) = 288 K$ $T_{2} = 273 K$
Since, $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
So, $V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$
$= \frac{761 \times 900 \times 273}{288 \times 760}$
$= 854.2 m L$ .

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Initial conditions	NTP conditions
$V_{1} = 900 m L$	$V_{2} = ?$
$P_{1} = 761 m m$	$P_{2} = 760 m m$
$T_{1} = (273 + 15) = 288 K$	$T_{2} = 273 K$

1.22 g of a gas measured over water at 15∘C and a pressure of 775 mm of mercury occupied 900 mL. Calculate the volume of dry gas at NTP. Vapor pressure of water at 15∘C is 14 mm.

Pressure of dry gas = Pressure of moist gas − Aqueous tension=775−14=761 mmInitial conditionsNTP conditionsV1=900 mLV2=?P1=761 mmP2=760 mmT1=(273+15)=288 KT2=273 KSince, P1V1T1=P2V2T2So, V2=P1V1T2T1P2=761×900×273288×760=854.2 mL.

$1.22 g$ of a gas measured over water at $15^{\circ} C$ and a pressure of $775 m m$ of mercury occupied $900 m L$ . Calculate the volume of dry gas at NTP. Vapor pressure of water at $15^{\circ} C$ is $14 m m$ .