Answer:
22.5
Explanation:
Vrms=√((3k_B T)/m)
Let V1 = Vrms and V2 = 2 Vrms
V1=√((3k_B T)/m) and V1^2=(3k_B T)/m
V2=2√((3k_B T)/m) V2^2=(12k_B T)/m
Number of molecules is given by dN = N f(v) dV
f(v)=(4 )/п (m/(3k_B T))^(3/2) V^2 e^(-(mv^2)/(2K_B T))
f(v1)=(4 )/п (m/(3k_B T))^(3/2) 〖V1〗^2 e^(-(m〖v1〗^2)/(2K_B T))
f(v2)=(4 )/п (m/(3k_B T))^(3/2) 〖V2〗^2 e^(-(m〖v2〗^2)/(2K_B T))
The ratio of the number of molecules; (dN(v1))/(dN(v2))=Nf(v1)dV/Nf(v2)dV=(f(v1))/(f(v2))
(f(v1))/(f(v2))=((4 )/п (m/(3k_B T))^(3/2) 〖V1〗^2 e^(-(m〖v1〗^2)/(2K_B T)))/((4 )/п (m/(3k_B T))^(3/2) 〖V2〗^2 e^(-(m〖v2〗^2)/(2K_B T)) )=(〖V1〗^2 e^(-(m〖v1〗^2)/(2K_B T)))/(〖V2〗^2 e^(-(m〖v2〗^2)/(2K_B T)) )
Putting the values of v1 and v2 into (f(v1))/(f(v2)), the equation above reduces to
(dN(V1))/(dN(V2))=(f(V1))/(f(v2))=1/4 e^(-3/2)/e^(-12/2)
(dN(V1))/(dN(V2))=22.5
