# Question #9b58c

##### 1 Answer

#### Explanation:

All you have to do here is use the **molar mass** of argon to determine how many *moles* you get in that **ideal gas law** equation to find the pressure of the gas.

So, you know that argon has a molar mass of **one mole** of argon has a mass of

Use this value to find the number of moles of argon present in your sample

#325color(red)(cancel(color(black)("g"))) * "1 mole Ar"/(39.9color(red)(cancel(color(black)("g")))) = "8.145 moles Ar"#

The *ideal gas law* equation looks like this

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

*universal gas constant*

**absolute temperature** of the gas

At this point, the most important thing to do is make sure that you know the **units** used in the expression of the universal gas constant.

In your case, you can use the value

#R = 8.31color(white)(a)("Pa" * "m"^3)/("mol" * "K")#

http://www.cpp.edu/~lllee/gasconstant.pdf

This will allow you to plug the values given to you in the ideal gas law equation *without* doing any unit conversions.

As you can see, the pressure of the gas will be expressed in *pascals*,

#PV = nRT implies P = (nRT)/V#

Plug in your values to get

#P = (8.145color(red)(cancel(color(black)("moles"))) * 8.31("Pa" * color(red)(cancel(color(black)("m"^3))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 298color(red)(cancel(color(black)("K"))))/(0.005color(red)(cancel(color(black)("m"^3))))#

#P = "4,034,023 Pa"#

Since you only have one **sig fig** for the volume of the gas, you can only have one sig fig for the resulting pressure

#P = color(green)(|bar(ul(color(white)(a/a)4 * 10^6"Pa"color(white)(a/a)|)))#