Compute the root-mean-square speed of a nitrogen molecule at 79.8°C. The molar mass of nitrogen molecules (N2) is 28.0*10-3 kg/mol. At what temperatures will the root-mean-square speed be (b) 1/2 times that value and (c) 2 times that value?

Vrms = 560.462 m/s

temperature is 88.2 K

temperature = 1411.2 k

Explanation:

Given data

temperature = 79.8°C = 79.8 + 273 = 352.8 K

(N2) = 28.0 * 10^-3 kg/mol

to find out

temperatures and 1/2 times that value and 2 times that value

solution

we know velocity rms formula that is

Vrms = √ (3RT/m)

here R = 8.31 and T = 352.8 and m = 28.0*10^-3

so Vrms = √ (3 (8.31) 352.8 / 28*10^-3)

Vrms = 560.462 m/s

and

for 1/2 Vrms

we know as V is proportional to square root of temperature

so it will be 1/2th if temperature is made 1/4th

so t/4 = 352.8 / 4 = 88.2 K

so temperature is 88.2 K

and

if it will be 2 times if temp is made 4 time

temperature = 4 (352.8)

temperature = 1411.2 k