31m/s is a typical highway speed for a car. At what temperature do the molecules of nitrogen gas have an rms speed of 31m/s?
T=[(x^2)*m]/3k
so
x=31m/s
m=N2=2*14=28g=.028kg
k=1.38e^-23..I'm keep getting 6.333651477e^10...which is incorrect. What am I missing?
Molecules of N2 do not have a mass of .028kg. Perhaps a mole of them have that mass, so divide by avagrado's number for the mass of each.
To find the temperature at which the molecules of nitrogen gas have an RMS (root mean square) speed of 31 m/s, you can use the formula T = (x^2 * m) / (3k), as you have correctly identified.
Let's plug in the values:
x = 31 m/s (RMS speed of the molecules)
m = 28 g = 0.028 kg (molar mass of nitrogen gas)
k = 1.38 × 10^(-23) J/K (Boltzmann constant)
Now, the error in your calculation seems to stem from a unit conversion mistake. The molar mass of nitrogen gas is given in grams (g), but it needs to be in kilograms (kg) for the equation. So let's convert the molar mass from grams to kilograms:
m = 28 g = 0.028 kg
Now, let's substitute the values into the equation:
T = (x^2 * m) / (3k)
T = (31^2 * 0.028) / (3 * 1.38 × 10^(-23))
T = (961 * 0.028) / (4.14 × 10^(-23))
T ≈ 0.066 K
Therefore, the temperature at which the molecules of nitrogen gas have an RMS speed of 31 m/s is approximately 0.066 K.