At 250 K, vrms = 750 m/s for a molecule of an ideal gas. The temperature of the gas is raised until vrms = 1100 m/s. What will the average translational kinetic energy of a molecule be at this temperature? ____ x 10^-20 J
Added by Steven B.
Step 1
We know that the root-mean-square velocity (vrms) of a molecule is related to its kinetic energy (KE) by the equation: vrms = √(3kT/m) where k is the Boltzmann constant, T is the temperature in Kelvin, and m is the mass of the molecule. Show more…
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