# Question #732f1

##### 1 Answer

No, you will not.

#### Explanation:

First of all, the units for the *ideal gas constant*,

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

That said, you can prove that this value of **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

*universal gas constant*

**absolute temperature** of the gas

Rearrange to solve for

#R = (PV)/(nT)#

Notice that the units given to you for

#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*,

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*,

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

Now, you can express the volume in *liters*, *kilopascals*,

#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 *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 *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

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