A #4.60*L# volume of gas at #845*mm*Hg# pressure is expanded such that the new pressure is #368*mm*Hg#. To what volume does it expand?

2 Answers
Sep 4, 2017

Answer:

A measurement of #845*mm*Hg# is illegitimate.......

Explanation:

See this old question.......

To solve this question we would use.....

#V_2=(P_1V_1)/P_2#, but we would insist on kosher units.......

Sep 4, 2017

Answer:

#V_2 = 10.6# #"L"#

Explanation:

NOTE: Ideally, measurements of pressures greater than #760# #"mm Hg"# are non-ideal, because mercury barometers only measure up to that value. The equivalent unit, the #"torr"#, should be used if the pressure value exceeds #760# #"mm Hg"#.

We're asked to find the volume necessary for a gas system to exert a pressure of #368# #"mm Hg"#, assuming no change in temperature or amount of gas.

To do this, we can use the pressure-volume relationship of gases illustrated by Boyle's law:

#ulbar(|stackrel(" ")(" "P_2V_1 = P_2V_2" ")|)" "# (constant temperature and quantity)

where

  • #P_1# and #P_2# are the initial and final pressures of the gas, respectively

  • #V_1# and #V_2# are the inital and final volumes of the gas, respectively

We know:

  • #P_1 = 845# #"mm Hg"#

  • #V_1 = 4.60# #"L"#

  • #P_2 = 368# #"mm Hg"#

  • #V_2 = ?#

Let's rearrange the equation to solve for the final volume, #V_2#:

#V_2 = (P_1V_1)/(P_2)#

Plugging in known values:

#color(red)(V_2) = ((845cancel("mm Hg"))(4.60color(white)(l)"L"))/(368cancel("mm Hg")) = color(red)(ulbar(|stackrel(" ")(" "10.6color(white)(l)"L"" ")|)#