Question

At what temperature is the root mean square speed of an atom in a Argon gas cylinder equals to the rms speed of a helium gas atom at -20° C? (atomic mass of Ar = 39.9 u, of He = 4.0 u)

Answer 1

at 18 degree Celsius

Answer 2

Hey mate,

◆ Answer- 2257 °C

◆ Explaination-
# Given-
T1 = –20 °C = 253 K
M1 = 4 AMU
M2 = 40 AMU (approx)

# Solution-
RMS velocity of helium is calculated by-
vrms(He) = √(3RT1/M1)

RMS velocity of argon is calculated by-
vrms(Ar) = √(3RT2/M2)

Given that, at T1 = 253 k
vrms(He) = vrms(Ar)
√(3RT1/M1) = √(3RT2/M2)

Rearranging,
T2 = T1M2 / M1
T2 = 253 × 40 / 4
T2 = 2530 K
T2 = 2257 °C

Therefore, the temperature of the argon atom will be 2257 °C.

Hope this is helpful...