U3.4 · Ideal Gas LawPractice · 理想气体定律

n = PV/RT = 3.50 × 5.00 ÷ (0.08206 × 298.15) = 0.7153 mol, and m = 0.7153 × 28.02 = 20.0 g. Convert the temperature to kelvins, and use R in L·atm/(mol·K) to match the units given.

The gas is a mixture of $\ce{H2}$ and water vapor, so by Dalton’s law, P($\ce{H2}$) = 99.80 − 2.645 = 97.16 kPa. n = PV/RT = 97.16 × 0.2500 ÷ (8.314 × 295.15) = $9.90 \times 10^{-3}$ mol. Using the total pressure would count the water vapor as hydrogen.

From PV = nRT and n = m/M, d = m/V = PM/RT = 1.00 × 44.01 ÷ (0.08206 × 323.15) = 1.66 g/L. The hotter the gas, the more spread out and the less dense it is; at 0 °C it would be about 1.96 g/L.

n(He) = 4.00 ÷ 4.00 = 1.00 mol and n($\ce{O2}$) = 16.0 ÷ 32.00 = 0.500 mol. Each gas’s partial pressure depends only on its own moles: P(He) = 1.00 × 0.08206 × 300 ÷ 10.0 = 2.46 atm. $\ce{O2}$ has the larger mass but a partial pressure of only 1.23 atm; the total is 3.69 atm.

With volume and moles fixed, pressure is proportional to absolute temperature: P₂ = 2.00 × 450.15 ÷ 300.15 = 3.00 atm. Using Celsius gives 2.00 × 177 ÷ 27 = 13.1 atm, which is wrong, and 1.33 atm inverts the ratio.