2) Estimate the effusion rate of methane through a small hole in a pipeline operating at 75 psia, if a leak test with helium at the same conditions reveals that \( 0.55 \mathrm{dm}^{3} \) of helium escapes per minute. Moreover, if the cost of gas per thousand cubic feet is \( 1.25 \mathrm{US} \$ \). Which would be the lost per year?

To estimate the effusion rate of methane, you can use Graham’s Law of Effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Since the molar mass of helium is about 4 g/mol and that of methane is about 16 g/mol, this implies that methane will effuse at a rate slower than helium. By applying Graham’s Law, you find that the effusion rate of methane would be \( 0.55 \mathrm{dm}^{3} \times \sqrt{\frac{4}{16}} = 0.275 \mathrm{dm}^{3} \) per minute. Now, let's put a price on that leakage. First, convert the volume of methane that escapes annually: \( 0.275 \mathrm{dm}^3 \times 60 \, \text{min/h} \times 24 \, \text{h/day} \times 365 \, \text{days} = 144,825 \, \mathrm{dm}^3 \) or approximately \( 144.83 \, \text{m}^3 \) per year. Since \( 1 \, \mathrm{m}^3 \) equals about \( 35.3147 \, \text{ft}^3 \), you get around \( 5,128 \, \text{ft}^3 \). With gas costing $1.25 per 1,000 ft³, the annual loss would amount to \( 1.25 \times 5.128 \approx \$6.41 \) per year, a small but needless expense!