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 - the universal gas constant
T - the absolute temperature of 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
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,
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.