# 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?

##### 1 Answer

I got

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

#\mathbf(upsilon_("RMS") = sqrt((3RT)/("M"_m))# where:

#R# is theuniversal gas constant,#"8.314472 J/mol"cdot"K"# , where#"1 J" = "1 kg"cdot"m"^2"/s"^2# .#T# is thetemperaturein#"K"# .#"M"_m# is themolar massof the gas in#"kg/mol"# (NOT#"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:

#PV = nRT#

Since you were given the density,

#color(green)((PV)/n = RT)#

Now you can substitute into the RMS-speed equation.

#upsilon_("RMS") = sqrt((3PV)/(nM_m))#

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

#V/(nM_m) stackrel(?)(=) 1/rho# #larr# reciprocal density! AHA!

The left side has units of

So, what we have in the end is:

#color(blue)(upsilon_("RMS")) = sqrt((3RT)/(M_m))#

#= sqrt((3PV)/(nM_m))#

#= sqrt((3P)/(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(3/4*(1.2xx10^5 "m"^2"/s"^2))#

#=# #color(blue)("300 m/s")#