33. If \( 140 \mathrm{~cm}^{3} \) of methane diffuse into air in 72 seconds hew leng will it take \( 210 \mathrm{~cm}^{3} \) of susphut dioride to diffuse inder the same conditions? 34. The temprature of \( 2 s ~ L \) of a qas inctially at sip is raised to \( 250^{\circ} \mathrm{C} \) at constant volisme: calculate the final pressure. of the gas is atmosphere.

Below are the step‐by‐step solutions for each problem. ────────────────────────────── Problem 33 We are told that 140 cm³ of methane diffuses in 72 seconds. We wish to know how long 210 cm³ of sulfur dioxide will take to diffuse under the same conditions. Recall from Graham’s law that the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its molar mass. Since we are dealing with rates measured in volume per time under identical conditions, the diffusion rates (r) obey   r ∝ 1/√M The diffusion “rate” is essentially volume/time. Let the rates for methane and sulfur dioxide be r₁ and r₂, respectively. Then   r₁/r₂ = √(M₂/M₁) Since rate r = (volume/time), we can write   (V₁/t₁) / (V₂/t₂) = √(M₂/M₁) We now identify the given quantities: • For methane, V₁ = 140 cm³, t₁ = 72 s, and M₁ = 16 g/mol (CH₄: 12 + 4 = 16). • For sulfur dioxide, V₂ = 210 cm³, t₂ = ? and M₂ = 64 g/mol (SO₂: 32 + 2×16 = 64). Rearrange the above relation to solve for t₂:   t₂ = t₁ × (V₂/V₁) × √(M₂/M₁) Plug in the numbers:   t₂ = 72 s × (210/140) × √(64/16) Calculate step‐by‐step: 1. Volume ratio: 210/140 = 1.5. 2. Molar mass ratio inside the square root: 64/16 = 4, so √4 = 2. Thus,   t₂ = 72 × 1.5 × 2 = 216 seconds ────────────────────────────── Problem 34 We are told that the temperature of a given volume of gas is raised at constant volume and that the initial pressure is 1 atmosphere. Because the volume is fixed, the pressure of an ideal gas is directly proportional to its absolute temperature (in Kelvin). This relationship is given by   P/T = constant   or   P₁/T₁ = P₂/T₂ Here, the initial pressure P₁ is 1 atm. It is stated that the gas is “initially at sip” – this may be interpreted as being at standard conditions. At standard conditions the temperature is usually taken as 0°C, which is 273 K. The final temperature is given as 250°C which must be converted to Kelvin:   T₂ = 250 + 273 = 523 K Now, applying the proportionality:   P₂ = P₁ × (T₂/T₁) = 1 atm × (523/273) Compute the ratio:   523/273 ≈ 1.916 Thus, the final pressure is approximately 1.92 atmospheres. ────────────────────────────── Answers 33. It will take about 216 seconds for 210 cm³ of sulfur dioxide to diffuse. 34. The final pressure of the gas, when raised to 250°C at constant volume, is approximately 1.92 atm.