Exactly 100 $m L$ of oxygen is collected over the water of $23^{\circ} C$ and 800 $m m$ pressure. Calculate the approximate value of the volume of dry oxygen at $N T P$ . (as nearest integer)
[Vapour pressure of water at $23^{\circ} C$ is 21 $m m$ .]

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Solution

$V_{1} = 100 m L$
$V_{2} = ?$
$P_{1} = 779 m m$
$P_{2} = 760 m m$
$T_{1} = 23^{\circ} C = 23 + 273$
$T_{2} = 273 K$
$= 296 K$

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

$V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}} = \frac{779 \times 100 \times 273}{760 \times 296} = 94.54 m L$

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$V_{1} = 100 m L$	$V_{2} = ?$
$P_{1} = 779 m m$	$P_{2} = 760 m m$
$T_{1} = 23^{\circ} C = 23 + 273$	$T_{2} = 273 K$
$= 296 K$

Exactly 100 mL of oxygen is collected over the water of 23∘C and 800 mm pressure. Calculate the approximate value of the volume of dry oxygen at NTP. (as nearest integer)[Vapour pressure of water at 23∘C is 21 mm.]

V1=100mLV2=?P1=779mmP2=760mmT1=23∘C=23+273 T2=273K =296KP1V1T1=P2V2T2V2=P1V1T2P2T1=779×100×273760×296=94.54mL

Exactly 100 $m L$ of oxygen is collected over the water of $23^{\circ} C$ and 800 $m m$ pressure. Calculate the approximate value of the volume of dry oxygen at $N T P$ . (as nearest integer)
[Vapour pressure of water at $23^{\circ} C$ is 21 $m m$ .]

$V_{1} = 100 m L$
$V_{2} = ?$
$P_{1} = 779 m m$
$P_{2} = 760 m m$
$T_{1} = 23^{\circ} C = 23 + 273$
$T_{2} = 273 K$
$= 296 K$

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

$V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}} = \frac{779 \times 100 \times 273}{760 \times 296} = 94.54 m L$