# At What Temperature is the Root Mean Square Speed of an Atom in an Argon Gas Cylinder Equal to the Rms Speed of a Helium Gas Atom at – 20 °C? (Atomic Mass of Ar = 39.9 U, of He = 4.0 U). - Physics

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

#### Solution 1

Temperature of the helium atom, THe = –20°C= 253 K

Atomic mass of argon, MAr = 39.9 u

Atomic mass of helium, MHe = 4.0 u

Let, (vrms)Ar be the rms speed of argon.

Let (vrms)He be the rms speed of helium.

The rms speed of argon is given by:

(v_"rms")_"Ar" = sqrt((3RT_"Ar")/M_"Ar") .... (1)

Where,

R is the universal gas constant

TAr is temperature of argon gas

The rms speed of helium is given by:

(v_"rms")_"He" = sqrt((3RT_"He")/M_"He") ... (ii)

It is given that

(vrms)Ar = (vrms)He

sqrt((3RT_"Ar")/ M_"Ar") = sqrt((3RT_"He")/M_"He")

T_"Ar"/M_"Ar" = T_"He"/M_"He"

T_"Ar" = T_"He"/M_"He"  xx M_"Ar"

= 253/4 xx 39.9

= 2523.675 = 2.52 × 103 K

Therefore, the temperature of the argon atom is 2.52 × 103 K.

#### Solution 2

Let C and C’ be the rms velocity of argon and a helium gas atoms at temperature T K and T K respectively

Here, M = 39.9; M’ = 4.0; T =?; T = -20 + 273 = 253 K

Now, C = sqrt((3RT)/M
)= sqrt((3RT)/39.9) and C' = sqrt((3RT')/M') = sqrt(3R xx 253)/4

Since C =C'

Therefore sqrt((3RT)/39.9) =  sqrt(3Rxx253)/4 or T = (39.9 xx 253)/4  = 2523.7 K

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 13 Kinetic Theory
Q 9 | Page 334