(b) Calculate the partial pressure of helium gas
Assuming that each alpha particle becomes a helium atom, we can calculate the number of moles of helium atoms formed using Avogadro's number:
\(number \,of\,moles = \frac{number\,of\,atoms}{Avogadro's\,number}\)
\(number\,of\,moles\,of\,helium = \frac{3804240}{6.022\times10^{23}\,atoms/mol} = 6.318\times10^{-19}\,mol\)
Now, we can use the ideal gas law formula (PV = nRT) to find the partial pressure of helium gas. But first, we need to convert the temperature given in Celsius to Kelvin.
Temperature in Kelvin: \(T(K) = T(^\circ C) + 273.15 = 22^\circ C + 273.15 = 295.15\,K\)
The volume of the container is given as 25.0 mL, which we need to convert to L.
Volume in L: \(V(L) = 25.0\,mL \times \frac{1\,L}{1000\,mL} = 0.025\,L\)
And we need to use the gas constant R in appropriate units, which is: \(R = 0.0821\,atm\,L/mol\,K\)
Now using the ideal gas law formula and solving for pressure, we have:
\(P = \frac{nRT}{V}\)
\(P = \frac{(6.318\times10^{-19}\,mol)(0.0821\,atm\,L/mol\,K)(295.15\,K)}{0.025\,L}\)
\(P = 5.20\times10^{-17}\,atm\)
So, the partial pressure of helium gas in the container after 245 days is 5.20 x 10⁻¹⁷ atm.