# If the density of the gas is "4 kg/m"^3 and its pressure is 1.2 xx 10^5 "N/m"^2, how do I calculate the root-mean-square speed?

May 13, 2016

I got $\text{300 m/s}$.

The equation for the root-mean-square speed is:

$\setminus m a t h b f \left({\upsilon}_{{\text{RMS") = sqrt((3RT)/("M}}_{m}}\right)$

where:

• $R$ is the universal gas constant, $\text{8.314472 J/mol"cdot"K}$, where ${\text{1 J" = "1 kg"cdot"m"^2"/s}}^{2}$.
• $T$ is the temperature in $\text{K}$.
• ${\text{M}}_{m}$ is the molar mass of the gas in $\text{kg/mol}$ (NOT $\text{g/mol}$!).

Given that you were not given any identity for the gas, this question is probably either assuming ideality, or somehow the variables cancel so you don't need the molar mass.

Let's say we considered the ideal gas law:

$P V = n R T$

Since you were given the density, $\rho = {\text{4 kg/m}}^{3}$, and a pressure, $1.2 \times {10}^{5} {\text{N/m}}^{2}$ (or $\text{Pa}$), here's one way you could do it.

$\textcolor{g r e e n}{\frac{P V}{n} = R T}$

Now you can substitute into the RMS-speed equation.

${\upsilon}_{\text{RMS}} = \sqrt{\frac{3 P V}{n {M}_{m}}}$

...But wait! Let's consider this chunk of the equation for a minute.

V/(nM_m) stackrel(?)(=) 1/rho $\leftarrow$ reciprocal density! AHA!

The left side has units of "L"/("mol"cdot"kg/mol"), which cancels out to give $\text{L"/"kg}$ (i.e. the units of the reciprocal density)!

So, what we have in the end is:

$\textcolor{b l u e}{{\upsilon}_{\text{RMS}}} = \sqrt{\frac{3 R T}{{M}_{m}}}$

$= \sqrt{\frac{3 P V}{n {M}_{m}}}$

$= \sqrt{\frac{3 P}{\rho}}$

= sqrt((3*(1.2xx10^5 "N")/"m"^2)/(("4 kg")/"m"^3))

= sqrt((3*(1.2xx10^5 "N"))/cancel("m"^2)*("1 m"^(cancel(3)^(1)))/("4 kg"))

= sqrt((3*(1.2xx10^5 "N"cdot"m"))/("4 kg"))

= sqrt((3*(1.2xx10^5 cancel"kg"cdot"m"^2"/s"^2))/("4" cancel"kg"))

$= \sqrt{\frac{3}{4} \cdot \left(1.2 \times {10}^{5} {\text{m"^2"/s}}^{2}\right)}$

$=$ $\textcolor{b l u e}{\text{300 m/s}}$