8
Problem solving
Textbook: pp. 28–29
GoalSolve problems on the speed, energy and temperature of gas molecules.
New words
root-mean-square speed · o‘rtacha kvadratik tezlikaverage kinetic energy · o‘rtacha kinetik energiyaspeed ratio · tezlik nisbatirounding · yaxlitlash
Explanation
Three formulas underlie the problems of this topic: E_k = (3/2)kT, v = √(3RT/M) and p = nkT. First decide what is asked, choose the right formula and express the unknown (e.g. T = Mv²/(3R)). Temperature is always in kelvins and molar mass in kg/mol. When comparing two states the constants cancel: if M and R are the same, v₂/v₁ = √(T₂/T₁). Check that the result makes sense: when temperature rises the speed must rise, and in a heavier gas the speed is smaller.
Worked examples
At what temperature do helium atoms (M = 4 g/mol) have the same root-mean-square speed as methane molecules (M = 16 g/mol) at 400 K? v ∝ √(T/M), so T_He/M_He = T_CH₄/M_CH₄ and T_He = 400·4/16 = 100 K.
A gas at 400 K is cooled until the root-mean-square speed of its molecules halves. Since v ∝ √T, T falls 4 times: T₂ = 400/4 = 100 K, so the temperature dropped by 300 K.
Class activity
Two-team contest: the teacher gives a gas and a temperature (e.g. nitrogen, 300 K); teams calculate v and sanity-check the answer; the team with a correct, reasoned answer scores. Calculators may be used.
Practice
1
Estimate the root-mean-square speed of oxygen molecules (M = 0.032 kg/mol) at 27 °C.
about 4.8·10² m/s
2
Find the temperature of a gas whose molecules have average kinetic energy 8.28·10⁻²¹ J (k = 1.38·10⁻²³ J/K).
400 K
3
A gas is at 250 K. By how many kelvins must its temperature be raised to double the root-mean-square speed of its molecules?
750
4
Why does light hydrogen escape from Earth’s atmosphere more easily than heavy gases?
At the same temperature hydrogen molecules move faster, so many more of them reach escape speed.