Question #732f1

1 Answer
Jun 3, 2016

Answer:

No, you will not.

Explanation:

First of all, the units for the ideal gas constant, #R#, that you are working with are actually

#8.314 "J"/("mol" * "K")#

That said, you can prove that this value of #R# will not give you a correct answer if the pressure is expressed in mmHg by using the ideal gas law

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#

where

#P# - the pressure of the gas
#V# - the volume it occupies
#n# - the number of moles of gas
#R# - the universal gas constant
#T# - the absolute temperature of the gas

Rearrange to solve for #R#

#R = (PV)/(nT)#

Notice that the units given to you for #R# are

#R = ["J"/("mol" * "K")]#

Compare this with the above equation to see that the units for pressure and volume must multiply to give you joules, #"J"#.

Since you know that

#"1 J" = "1 N" xx "1 m"#

you can say that the product of the pressure and the volume will be expressed in

#P * V = "J" = "N" * "m"#

Now, if you were to express the volume in cubic meters and the pressure in pascals, #"Pa"#, you would have

#P * V = "N" * "m" = overbrace("N"/"m"^2)^(color(blue)("= Pa")) * "m"^3 = "Pa" * "m"^3#

Now, you can express the volume in liters, #"L"#, and the pressure in kilopascals, #"kPa"#

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 L" = 10^(-3)"m"^3)color(white)(a/a)|)))" "# and #" "color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kPa" = 10^(3)"kPa")color(white)(a/a)|)))#

to get

#P * V = color(red)(cancel(color(black)(10^3)))"kPa" * color(red)(cancel(color(black)(10^(-3))))"L" = "kPa" * "L"#

Therefore, in order to have

#R = 8.314 "J"/("mol" * "K")#

you need to have

#R = 8.314("kPa" * "L")/("mol" * "K")#

This means that using this value of #R# when the pressure is expressed in mmHg will not get you a correct answer.

In fact, you can use the following conversion factors

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = " 760 mmHg")color(white)(a/a)|)))" "# and #" "color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = "101.325 kPa")color(white)(a/a)|)))#

to find the correct value of #R# for when the pressure is expressed in mmHg.

You will have

#8.314 (color(red)(cancel(color(black)("kPa"))) * "L")/("mol" * "K") * (1color(red)(cancel(color(black)("atm"))))/(101.325color(red)(cancel(color(black)("kPa")))) * "760 mmHg"/(1color(red)(cancel(color(black)("atm")))) = 62.36("mmHg" * "L")/("mol" * "K")#

Therefore, the correct value of #R# to use in this case is

#color(green)(|bar(ul(color(white)(a/a)color(black)(R = 62.36("mmHg" * "L")/("mol" * "K"))color(white)(a/a)|)))#