Actinium-227 has a half life of $8 x 10^{3}$ $8 x 10^3$ days, decaying by a $α$ $alpha$ -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?

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Actinium-227 has a half life of $8 x 10^{3}$ $8 x 10^3$ days, decaying by a $α$ $alpha$ -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?

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Answer 1 · 2017-01-12T23:22:14

$V \approx 0.1 L$ $V ~~ 0.1" L"$

Explanation:

Covert $8 \times 10^{3}$ $8xx10^3$ days to years

$\frac{8000 days}{1} \frac{1 year}{365.25 days} = 21.9 years$ $(8000" days")/1(1" year")/(365.25" days") = 21.9" years"$

Half life equation is:

$\frac{1}{2} = e^{γ t}$ $1/2 = e^(gammat)$

Substitute 21.9 years for t:

$\frac{1}{2} = e^{γ (21.9 years)}$ $1/2 = e^(gamma(21.9" years"))$

Solve for $γ$ $gamma$

$ln (\frac{1}{2}) = γ (21.9 years)$ $ln(1/2) = gamma(21.9" years")$

$γ = - \frac{ln (2)}{21.9 years}$ $gamma = -ln(2)/(21.9" years")$

$Q (t) = Q_{0} e^{- ln (2) \frac{t}{21.9 y e a r s}}$ $Q(t) = Q_0e^(-ln(2)t/(21.9" years))$

Starting with 1 gram:

$Q (t) = e^{- ln (2) \frac{t}{21.9 y e a r s}}$ $Q(t) = e^(-ln(2)t/(21.9" years))$

Evaluate at t = 100 years:

$Q (t) = (1 g) e^{- ln (2) \frac{100 years}{21.9 y e a r s}}$ $Q(t) = (1" g")e^(-ln(2)(100 " years")/(21.9" years))$

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

$\frac{1 g}{227 g/mol} (1 - e^{- ln (2) \frac{100 years}{21.9 y e a r s}})$ $(1" g")/(227" g/mol")(1 - e^(-ln(2)(100 " years")/(21.9" years)))$

$4.2 \times 10^{- 3} mol$ $4.2xx10^-3" mol"$ has decayed creating the same number of mole of helium nuclei:

$n = 4.2 \times 10^{- 3} mol$ $n = 4.2xx10^-3" mol"$

Convert the temperature to Kelvin:

$T = 25 + 273 = 298 K$ $T = 25 + 273 = 298 K$

Convert mmHg to kPa:

$P = \frac{748 mmHg}{1} \frac{0.1333 kPa}{1 mmHg} = 99.7 kPa$ $P = (748" mmHg")/1(0.1333" kPa")/(1" mmHg") = 99.7" kPa"$

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

$V = \frac{n R T}{P}$ $V = (nRT)/P$

$V \approx 0.1 L$ $V ~~ 0.1" L"$

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Actinium-227 has a half life of $8 x 10^{3}$ $8 x 10^3$ days, decaying by a $α$ $alpha$ -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?

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Explanation:

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Actinium-227 has a half life of 8x1038 x 10^3 days, decaying by a αalpha-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

Explanation:

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Actinium-227 has a half life of $8 x 10^{3}$ $8 x 10^3$ days, decaying by a $α$ $alpha$ -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?