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).

Question

Philoid · Accepted Answer

All the three vessels have the same capacity, which implies they all have the same volume. Thus, each gas has the same pressure, volume, and temperature. According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules.

Given:

Temperature of the helium atom(T_h)

T_h = –20°C = 253 K

Atomic mass of argon(M_a) = 39.9 u

Atomic mass of helium(M_h) = 4u

Let, V_a be the rms speed of argon and let V_h be the rms speed of helium.

Since, MV² = k_BT (let k_B be simply k)

∴ V =

The rms speed of argon is given by-

V_a = ---(1)

The rms speed of helium is given by-

V_h = ---(2)

∴ since V_a = V_h

⇒ =

⇒ T_a = M_a

⇒ T_a = × 39.9 u

⇒ T_a = 2.52 × 10³ K

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