A sealed rigid vessel contains an ideal gas with molar mass 0.0280 kg mol⁻¹ at 300 K and pressure 1.150 × 10^5 Pa. Use R = 8.31 J mol⁻¹ K⁻¹. From pV = nRT and p = ρv_rms²/3, determine the gas density and root-mean-square molecular speed. The absolute temperature is then doubled without changing the amount of gas or vessel volume. Predict the new pressure, density and rms speed; explain why the average velocity vector can still be zero.

Question

Accepted Answer

Density 1.292 kg m⁻³; rms speed 516.8 m s⁻¹. After heating: pressure 2.300 × 10^5 Pa, unchanged density, rms speed 730.9 m s⁻¹. The number of moles is n = m/M. Inserting this into pV = nRT gives p = (m/V)RT/M = ρRT/M, so the molecular mass scale must be expressed in kilograms per mole. Rearrange to ρ = pM/(RT) = 1.292 kg m⁻³. The kinetic-theory equation then gives v_rms = √(3p/ρ) = √(3RT/M) = 516.8 m s⁻¹. At fixed amount and volume, pressure is proportional to absolute temperature. Doubling T gives p_new = 2.300 × 10^5 Pa; density stays fixed because neither mass nor volume changes. The rms speed scales as √T, not T, so v_rms,new = √2 v_rms = 730.9 m s⁻¹. Substitution into p = ρv_rms²/3 confirms the doubled pressure. Molecules move randomly in all directions, so positive and negative velocity components cancel in the average vector. Squaring speeds before averaging prevents this cancellation, giving a nonzero rms speed even with zero bulk flow.

Molecular speed and gas pressure · set 3

What a complete response should do

How did this one feel?