2.30 L of air at 4.53 atm is expanded to 2219 mmHg. What is the final volume in mL?

2 Answers
Feb 13, 2017

#2219# #mm*Hg# is an absurd unit.

Explanation:

#1*atm# will support a column of mercury #760*mm# high. Often, we quote the pressure as #760*mm# #Hg# (or thereabouts) in order to report daily fluctuations in pressure. In fact due to safety concerns, mercury has almost disappeared from modern laboratories.

It is tempting to say that #2219# #mm*Hg#

#-=# #(2219*mm*Hg)/(760*mm*Hg*atm^-1)~=3*atm#

But I would resist this temptation.

Feb 13, 2017

The final volume will be #"3570 mL"#, rounded to three significant figures.

Explanation:

First the units of pressure and volume need to be the same.

Pressure

#"1 atm=760.0 mmHg"#

Convert mmHg to atm.

#2219color(red)(cancelcolor(black)("mmHg"))xx(1"atm")/(760.0color(red)(cancelcolor(black)("mmHg")))="2.920 atm"#

Volume

#"1 L=1000 mL"#

Convert #"L"# to #"mL"#.

#2.30color(red)(cancel(color(black)("L")))xx(1000"mL")/(1color(red)(cancel(color(black)("L"))))="2300 mL"=2.30xx10^3"mL"#

The number of mL above is given to three significant figures using scientific notation. I will use #"2300 mL"# for convenience only.

This question is concerns Boyle's Law , which states that the volume #(V)# of a gas held at constant amount and temperature, is inversely proportional to the pressure #(P)#. This means that if the pressure goes up, the volume goes down, and vice-versa. The equation to use is:

#P_1V_1=P_2V_2#

Write what you know:

#P_1="4.53 atm"#
#V_1=2.30xx10^3"mL"#
#P_2="2.920 atm"#

Write what you don't know: #V_2#.

Solution
Rearrange the equation to isolate #V_2#. Substitute the known quantities into the equation and solve.

#V_2=(P_1V_1)/(P_2)#

#V_2=(4.53color(red)(cancel(color(black)("atm")))xx2300"mL")/(2.920color(red)(cancel(color(black)("atm"))))="3570 mL"# rounded to three significant figures