For a face on Galaxy, the width of the spectral line is due to the Brownian motion of the gas in the galaxy.

½ m v^{2} = 3/2 k T

where m = mass of the hydrogen atom = 2 x 10^{-24} grams and k =
Boltzman's constant = 1.4 x 10^{-16 }ergs / Kelvin. Thus:

T = m v^{2} / 3 k

or:

T = 4.8 x 10^{-9 }v^{2}

Example:

For NGC628, the line width = 40 km/s = 4 x 10

^{6 }cm/sThen, T = 76, 000 K