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In a sample of hydrogen sulfide (M = 34.1 g/mol) at a temperature of 3.00 × 102 K, estimate the ratio of the number of molecules that have speeds very close to vrms to the number that have speeds very close to 2vrms.

Respuesta :

Answer:

The required ratio is 22.27.

Explanation:

Vr.m.s=speed of molecules

R=universal gas constant

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Ver imagen madihasaadmamoon

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

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