A vessel with a volume of 125 cm^3 holds tritium (^ 3 H , t_ 1/2 =...

Using the ideal gas equation, the number of moles of tritium gas is:

$n = \dfrac{PV}{RT} = \dfrac{500 \times 10^3 \times 125 \times 10^{-6}}{8.314 \times 300} \approx 0.025 \text{ mol}$

Since tritium gas (^{3}\text{H}_2$) is diatomic, the total number of radioactive tritium atoms is:<br><br>$N = 2nN_A = 2 \times 0.025 \times 6.022 \times 10^{23} \approx 3.01 \times 10^{22}$<br><br>The decay constant$\lambda$is:<br><br>$\lambda = \dfrac{\ln 2}{t_{1/2}} = \dfrac{0.693}{12.3 \times 365.25 \times 24 \times 3600} \approx 1.786 \times 10^{-9} \text{ s}^{-1}$<br><br>The activity$A$of the gas is:<br><br>$A = \lambda N = (1.786 \times 10^{-9}) \times (3.01 \times 10^{22}) \approx 5.38 \times 10^{13} \text{ Bq}$<br><br>Converting the activity to Curies ($1 \text{ Ci} = 3.7 \times 10^{10} \text{ Bq}$):<br><br>$A = \dfrac{5.38 \times 10^{13}}{3.7 \times 10^{10}} \approx 1454 \text{ Ci}$<br><br>The closest value among the given options is$1448 \text{ Ci}$.<br><br>Answer:$1448 \text{ Ci}