Question

Question #732f1

Answer 1

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)|)))`