An ideal gas has molecular mass 4.65 × 10⁻²⁶ kg and temperature 310 K in a rigid sealed vessel at pressure 1.00 × 10⁵ Pa. Take k = 1.38 × 10⁻²³ J/K. Use ½m⟨c²⟩ = 3kT/2. (a) Calculate its r.m.s. molecular speed. (b) The kelvin temperature increases by a factor 1.44; find the new r.m.s. speed and pressure. (c) Explain why average molecular velocity is still zero for gas at rest.

Question

Accepted Answer

Initial r.m.s. speed 525.4 m/s; final 630.4 m/s; final pressure 1.44 × 10⁵ Pa; random velocity vectors cancel. Cancel the factors of one-half in the energy equation to obtain ⟨c²⟩ = 3kT/m. Therefore crms = √[(3 × 1.38 × 10⁻²³ × 310)/(4.65 × 10⁻²⁶)] = 525.4 m/s, with the whole ratio inside the root. R.m.s. speed scales with √T, so increasing kelvin temperature by 1.44 multiplies the speed by √1.44 = 1.20. The new speed is 630.4 m/s rather than 1.44 times the original. The amount of gas and vessel volume remain fixed. From the ideal-gas relation, pressure is proportional to absolute temperature, so the new pressure is 1.44 × 1.00 × 10⁵ = 1.44 × 10⁵ Pa. Individual molecules move rapidly in random directions. In a gas with no bulk flow, opposite velocity components cancel in the vector average, yielding zero average velocity. Squared speeds are nonnegative and do not cancel, so their mean square and its square root remain nonzero.

Molecular speeds in a heated gas 4

What a complete response should do

How did this one feel?