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Q. A quantity of gas is collected in a graduated tube over mercury. The volume of the gas at 20$^\circ$C is 50.0 mL and the level of the mercury in the tube is 100 mm above the outside mercury level. The barometer reads 750 mm. Volume of the gas at STP is

States of Matter

Solution:

$T_1$ = 20 + 273 = 293 K, $T_2$ = 273 K
$V_1$ = 50 mL.
$P_1$ = 750-100 = 650 mm Hg; $P_2$ = 760 mm Hg
$\frac{P_1V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{50 \times 650 \times 273}{293 \times 760}$
= 39.8 mL