Question

A quantity of gas is collected in a graduated tube over the mercury. The volume of gas at ${18}^{o}C$ is $50.0mL$ and the level of mercury in the tube is $100mm$ above the outside mercury level The barometer reads $750$ torr. Hence, volume at STP is approximately:

• A. $22mL$
• B. $40mL$
• C. $20mL$
• D. $44mL$

Answer 1

Answer: (B)

$40mL$

$P_1=650torr$

$V_1=50ml$

$T_1=18C+273=291K$

$K_{P_2}=760Torr$

$V_2=?$

$T_2=273K$

Ideal gas law $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Rearranges to $\therefore V_2=\frac{\left(P_1{V}_1{T}_2\right)}{\left(P_2{T}_1\right)}$

$V_2=\left(\frac{650\left)\right(50\left)\right(273)}{\left(760\right)\left(291\right)}\right)$

$V_2=40.11ml$