Will a gas sample have a root-mean-square speed of #v_(RMS) = sqrt((3RT)/m)#?
Yes, as long as the gas is reasonably isotropic (meaning it travels in any direction with no preference over another; this is provided the gas sample is homogeneous), and the temperature is reasonably high such that the gases are classical particles.
Those are the fundamental assumptions for a derivation of the Maxwell-Boltzmann distribution. That can be seen here:
Note that your equation is incorrect.
Also, you cannot then invoke the ideal gas law to rewrite
Any homogeneous gas sample is reasonably isotropic, and so it would follow Maxwell-Boltzmann statistics as long as the temperature isn't too low.
When that is the case, the speed distribution is given by the Maxwell-Boltzmann distribution:
#f(upsilon)dupsilon = 4pi(m/(2pik_BT))^(3//2)upsilon^2e^(-mv^2//2k_BT)dupsilon#
To determine the RMS speed, which is what you have, one would evaluate the integral for the average squared speed and then take its square root:
#color(green)(upsilon_(RMS)) = sqrt(<< upsilon^2 >>) = (int_(0)^(oo) upsilon^2f(upsilon)dupsilon)^(1//2)#
#= color(green)(sqrt((3k_BT)/m) = sqrt((3RT)/M))#
That was done here:
And so, you should see that