Question

Suppose there are two identical gas cylinders. One contains the monatomic gas radon (Rn), and the other contains an equal mass of the monatomic gas xenon (Xe). The pressures in the cylinders are the same, but the temperatures are different. Determine the ratio KERn/KEXe of the average kinetic energy of an atom of radon to the average kinetic energy of an atom of xenon.

Answer 1

Given: The two cylinders have equal volume and the pressure of the gases is equal.

• Let the volume of each gas (equal to the volume of cylinder) be V

• Let the pressure of each gas be P

• Let the mass of each gas be M

• Let the number of moles of Radon be n1

• Let the number of moles of Xenon be n2

• Let the molar mass of Xenon be m1

• Let the molar mass of Xenon be m2

According to the ideal gas equation:PV=nRT

The number of moles can be expressed in terms of the molar mass of the gas and the total mass of the gas as:

n =M/m =Total Mass/ Mass of 1 Mole

For Radon,

PV = n₁RT_{1 =} M/m_{​​​​​​1}​​​ * RT_{​​​​​1}

For Xenon,

PV=n₂RT_​​​2 = M/m_{​​​​​​2}​​​​​ * RT_{​​​​​2}

for Radon; PV = n RT = MRT mn » for Xenon; me PV = nRT₂ = M R Tz2 On dividing ④ by ④ we get, m2 ma - The Average Kinetic Energy per molecule of an Ideal gas is given by; KE = 3 kJ Here; Ky Boltzmann Const, T Temperature of has. The of Avg. KiE Radon peu atom is :- K € = 3 KT1 Үч Хепатр — ҚЕ. – КТ.

Ratio of two Energies is in The Molar mass of Radon & Xenon ame :- mi= 222 g/mol m₂ = 131 g/mol % KE - 122 = 1.69 КЕ, 3.K.E (Rn) = 1.69 K.Elxe) 131