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

the answer is b

explanation:

answer: decibel

explanation: