# Question #ccbc8

##### 1 Answer

#### Explanation:

Your starting point here will be the **ideal gas law** equation

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

Here you have

#P# - the pressure of the gas

#V# - the volume it occupies

#n# - the number of moles of gas

#R# - theuniversal gas constant

#T# - theabsolute temperatureof the gas

Rearrange the above equation to solve for

#PV = nRT implies R = (PV)/(nT)#

Now, notice that the problem doesn't provide you with the **number of moles** of nitrogen gas present in your sample. This means that you're going to have to improvise a bit.

As you know, the *number of moles* present in a sample of a given compound is equal to the **mass** of the sample divided by the **molar mass** of the compound.

If you take

#n = m/M_M#

Plug this into the above equation to get

#R = (PV)/(m/M_M * T) = (PV)/(m * T) * M_M#

Now, the problem tells you that the gas has a **density** of *density* is defined by **mass** divided by **volume**.

You already used

#rho = m/V implies V/m = 1/(rho)#

Plug this into the equation to get

#R = P/T * 1/(rho) * M_M#

Nitrogen gas, **do not** forget to convert the temperature from *degrees Celsius* to *Kelvin*

#R = "1 atm"/((273.15 + 0)"K") * 1/(1.250 color(red)(cancel(color(black)("g"))) "dm"^(-3)) * 28.0134 color(red)(cancel(color(black)("g"))) "mol"^(-1)#

#color(green)(|bar(ul(color(white)(a/a)color(black)(R = 0.082 ("atm" * "dm"^3)/("mol" * "K"))color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**, but don't forget that you only have one sig fig for the temperature and pressure of the sample.

The actual value of

#R ~~ 0.0820575("atm" * "dm"^3)/("mol" * "K")#

so this is an excellent result.