2.8 · GasesPractice · 气体

1 atm = 101.325 kPa = 760 mmHg. 65.2 kPa × $\dfrac{1\ \text{atm}}{101.325\ \text{kPa}}$ = 0.643 atm, or about 489 mmHg: only about two-thirds of sea-level pressure, which is why newcomers feel short of breath there.

At constant volume, pressure is proportional to the Kelvin temperature. Convert first: T₁ = 20 + 273.15 = 293.15 K, T₂ = 150 + 273.15 = 423.15 K. $P_2 = P_1 \times \dfrac{T_2}{T_1}$ = 2.50 atm × 423.15 ÷ 293.15 = 3.61 atm. Using °C directly would give 18.8 atm, which is wrong. That’s why aerosol cans warn you to keep them away from heat.

Use the combined gas law, $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$, with Kelvin temperatures: T₁ = 20 + 273.15 = 293.15 K, T₂ = −40 + 273.15 = 233.15 K. V₂ = V₁ × $\dfrac{P_1}{P_2}$ × $\dfrac{T_2}{T_1}$ = 5.00 L × 1.00 ÷ 0.262 × 233.15 ÷ 293.15 = 15.2 L. The lower pressure makes the gas expand; the cold shrinks it a little, but the pressure effect is much larger.

At the same temperature and pressure, a gas’s volume depends only on the number of moles, not on which gas it is. Convert each to moles: 0.30 mol $\ce{O2}$: 0.30 mol; 6.4 g $\ce{SO2}$: 0.10 mol; $3.01 \times 10^{23}$ molecules of $\ce{NH3}$: 0.50 mol; 4.48 L of $\ce{CO2}$ at STP: 0.20 mol. $3.01 \times 10^{23}$ molecules is 0.50 mol, the most, so it takes up about 0.50 × 22.4 = 11.2 L.

From PV = nRT, n = $\dfrac{PV}{RT}$ = $\dfrac{150.\ \text{atm} \times 20.0\ \text{L}}{0.08206 \times 298.15\ \text{K}}$ = 122.6 mol. m = 122.6 mol × 32.00 g/mol = 3.92 × 10³ g = 3.92 kg. With P in atm and V in L, use R = 0.08206 L·atm/(mol·K).

Average kinetic energy depends only on temperature, so at the same temperature He and CH₄ have the same average kinetic energy. With equal ½mv², the lighter molecule moves faster: the ratio of root-mean-square speeds is v(He) : v(CH₄) = $\sqrt{M(\ce{CH4}) / M(\ce{He})}$ = $\sqrt{16.04 / 4.00}$ = 2.00, so He molecules move about twice as fast.

In a mixture, each gas behaves as if it had the whole container to itself: its partial pressure is its own nRT ÷ V, whatever else is present (so the 1.60 g of CH₄ isn’t needed). n(CO₂) = 4.40 ÷ 44.01 = 0.09998 mol, and P(CO₂) = 0.09998 × 0.08206 × 298.15 ÷ 5.00 = 0.489 atm.

n(H₂O₂) = 3.40 ÷ 34.02 = 0.09994 mol. The equation gives 1 mol O₂ per 2 mol H₂O₂, so n(O₂) = 0.04997 mol. At STP, V = 0.04997 mol × 22.4 L/mol = 1.12 L.

At the same temperature and pressure, volume ratios equal mole ratios, so the coefficients apply directly to volumes: 20 mL of CO uses 10 mL of O₂ and makes 20 mL of CO₂, leaving 20 − 10 = 10 mL of O₂. The gas afterward is 20 mL of CO₂ plus 10 mL of O₂, 30 mL in all. Counting only the CO₂ (20 mL) forgets the excess O₂.

Rates are inversely proportional to the square root of molar mass: $\dfrac{\text{rate}(\ce{NH3})}{\text{rate}(\ce{HCl})} = \sqrt{\dfrac{36.46}{17.03}}$ = 1.463. Both start at the same moment and reach the meeting point together, so the distances are in the ratio of the rates: NH₃ covers 1.463 ÷ (1 + 1.463) of the tube, 100 × 1.463 ÷ (1 + 1.463) = 59.4 cm. Lighter NH₃ moves faster, so the ring forms nearer the HCl end.