Actinium-227 has a half life of #8 x 10^3# days, decaying by a #alpha#-emission. Suppose that the helium gas originating from the alpha particles were collected. What volume of helium at 25°C and 748 mmHg could be obtained from 1.00 g after 100 years?

1 Answer
Jan 12, 2017

Answer:

#V ~~ 0.1" L"#

Explanation:

Covert #8xx10^3# days to years

#(8000" days")/1(1" year")/(365.25" days") = 21.9" years"#

Half life equation is:

#1/2 = e^(gammat)#

Substitute 21.9 years for t:

#1/2 = e^(gamma(21.9" years"))#

Solve for #gamma#

#ln(1/2) = gamma(21.9" years")#

#gamma = -ln(2)/(21.9" years")#

#Q(t) = Q_0e^(-ln(2)t/(21.9" years))#

Starting with 1 gram:

#Q(t) = e^(-ln(2)t/(21.9" years))#

Evaluate at t = 100 years:

#Q(t) = (1" g")e^(-ln(2)(100 " years")/(21.9" years))#

However we actually want the difference in moles (not grams)

#(1" g")/(227" g/mol")(1 - e^(-ln(2)(100 " years")/(21.9" years)))#

#4.2xx10^-3" mol"# has decayed creating the same number of mole of helium nuclei:

#n = 4.2xx10^-3" mol"#

Convert the temperature to Kelvin:

#T = 25 + 273 = 298 K#

Convert mmHg to kPa:

#P = (748" mmHg")/1(0.1333" kPa")/(1" mmHg") = 99.7" kPa"#

#R = 8.134" L kPa/K mol"#

#V = (nRT)/P#

#V ~~ 0.1" L"#