Q1(b). Most probable speed from the Maxwell distribution

1. Function to maximise. The fraction of molecules per unit speed is f(c)=1/N(dN_c)/(dc)=4π((m)/(2π k_BT))^(3/2)c²e^-mc²/2k_BT. The most probable speed is the c at which f(c) is a maximum. The constant factor does not change the position of the maximum, so only g(c)=c²e^-mc²/2k_BT is differentiated.
2. Differentiate: (dg)/(dc)=2c e^-mc²/2k_BT+c²(-(mc)/(k_BT))e^-mc²/2k_BT=c e^-mc²/2k_BT(2-mc²k_BT)
3. Set dg/dc=0. c=0 and c→∞ give the minima (f=0). The maximum is at 2-mc²k_BT=0: cₘₚ=√((2k_BT)/(m))=√((2RT)/(M)) (R=N_Ak_B, M=N_Am)
4. It is a maximum: dg/dc>0 for c<cₘₚ (the c² rise) and dg/dc<0 for c>cₘₚ (the exponential fall wins).
5. Example, N₂ at 300 K (m=4.653×10⁻²⁶ kg): cₘₚ=2(1.3805×10⁻²³)(300)/4.653×10⁻²⁶=422 m s⁻¹; c=√(8k_BT/π m)=476 m s⁻¹; cᵣₘₛ=√(3k_BT/m)=517 m s⁻¹.
6. cₘₚ∝√(T/m): the peak moves to higher speed and becomes lower and broader as T rises; a heavier gas has its peak at a lower speed.

Application. the high-speed tail above cₘₚ holds the molecules with enough energy to react, which is why reaction rates rise steeply with temperature; c∝1/√M is the basis of Graham's law of effusion.