Question 732f1

Jun 3, 2016

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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

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 $R$

$R = \frac{P V}{n T}$

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, $\text{J}$.

Since you know that

$\text{1 J" = "1 N" xx "1 m}$

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

$P \cdot V = \text{J" = "N" * "m}$

Now, if you were to express the volume in cubic meters and the pressure in pascals, $\text{Pa}$, you would have

$P \cdot V = {\text{N" * "m" = overbrace("N"/"m"^2)^(color(blue)("= Pa")) * "m"^3 = "Pa" * "m}}^{3}$

Now, you can express the volume in liters, $\text{L}$, and the pressure in kilopascals, $\text{kPa}$

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

$P \cdot V = \textcolor{red}{\cancel{\textcolor{b l a c k}{{10}^{3}}}} \text{kPa" * color(red)(cancel(color(black)(10^(-3))))"L" = "kPa" * "L}$

Therefore, in order to have

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

you need to have

$R = 8.314 \left(\text{kPa" * "L")/("mol" * "K}\right)$

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

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{R = 62.36 \left(\text{mmHg" * "L")/("mol" * "K}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$