Question

A container filled with helium gas contains 1.4 × 10^23 atoms. The gas is initially held at a temperature of 0℃. If 88 J of heat is added to the gas, a) what is the new temperature? b) by what factor does vrms, change?

Answer 1

a) number of atoms in 1 mole = 6.023*10^23

number of moles in 1.4*10^23 atoms = 1.4*10^23 / 6.023*10^23 = 0.232

Q = nCp*dT ( Cp for monoatomic gas is 5R/2)

88 = 0.232*5*8.314*(T - 0) / 2

T = 18.21 ^oC

b) V(rms) = sqrt(3RT/M)

V(new ) /V(old) = sqrt(T(new) / T(old) )

V(new) / V(old ) = sqrt((273 + 18.21) / 273)

V(new) / V(old) = 1.03

V(new) = 1.03 times the V(old)