A pressure of #1.0# #atm# will support a #760# #mm# column of mercury. What atmospheric pressure is required to support a column #882# #mm# tall?

2 Answers
Jun 9, 2017

A pressure of #1.0# atmosphere (#atm#) supports a column of #760# #mm# of mercury.

Just using proportional reasoning, the pressure we are asked for will be: #P = 882/760 xx 1.0 = 1.16# #atm#

Jun 9, 2017

Answer:

#x=1.16# atm's to 2 decimal places #-># approximate answer

#x=882/760 -> 1 61/32 # atm's #-># exact answer

Explanation:

#color(blue)("Preamble")#

#color(brown)(ul("For multiply or divide:"))#
To move a value to the other side of the equals change it into 1. As multiply by 1 does not change the other value.

#color(brown)(ul("For add or subtract:"))#
To move a value to the other side of the equals change it to 0. As adding 0 does not change the other value.

#color(brown)("The shortcut methods use the above but skip steps")#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#

Let the unknown count in atm's be #x#

Known: 1 atm #-=760 #mmHg

Using ratio:

#("atm")/("mmHg") ->1/760-=x/882# where #-=# means 'equivalent to

To get #x# on its own multiply both sides by 882.

#color(green)(1/760color(red)(xx882)" "=" "x/882color(red)(xx882))#

#color(green)((1color(red)(xx882))/760" "=" "x xx(color(red)(882))/882 )#

#1.160526...." "=" "x#

#x=1.16# atm's to 2 decimal places #-># approximate answer

#x=882/760 -> 1 61/32 # atm's #-># exact answer