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NCERT Exemplar · Q7

Q.A vessel of volume VV contains a mixture of 1 mole of Hydrogen and 1 mole of Oxygen (both considered as ideal). Let f1(v)dvf_1(v)dv, denote the fraction of molecules with speed between vv and (v+dv)(v + dv) with f2(v)dvf_2(v)dv, similarly for oxygen. Then

(a) f1(v)+f2(v)=f(v)f_1(v) + f_2(v) = f(v) obeys the Maxwell's distribution law.
(b) f1(v)f_1(v), f2(v)f_2(v) will obey the Maxwell's distribution law separately.
(c) Neither f1(v)f_1(v), nor f2(v)f_2(v) will obey the Maxwell's distribution law.
(d) f2(v)f_2(v) and f1(v)f_1(v) will be the same.
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The Maxwell-Boltzmann distribution applies separately to each species in a mixture, so each gas independently follows the same functional form — but with its own mass. Hence f1(v)f_1(v) and f2(v)f_2(v) each obey the Maxwell distribution law, while their sum does not.

  1. The core idea: each gas has its own distribution. The Maxwell-Boltzmann speed distribution for an ideal gas at temperature TT is

f(v) dv=4π(m2πkT)3/2v2e−mv2/(2kT) dv.f(v) \, dv = 4\pi \left( \frac{m}{2\pi kT} \right)^{3/2} v^2 e^{-mv^2/(2kT)} \, dv.

This form depends on the molecular mass mm. Hydrogen (m1m_1) and oxygen (m2m_2) have different masses — m2≈16 m1m_2 \approx 16\,m_1 — so their distributions are not the same function of vv.

  1. Why f1(v)f_1(v) and f2(v)f_2(v) each obey Maxwell’s law.

    In a mixture of non-interacting ideal gases at thermal equilibrium, each species independently follows the Maxwell-Boltzmann distribution at the common temperature TT. There is no cross-coupling: collisions between unlike molecules merely exchange energy, but the equilibrium velocity distribution for each species is still the one that maximises its own entropy. So f1(v)f_1(v) is Maxwellian with mass m1m_1, and f2(v)f_2(v) is Maxwellian with mass m2m_2.

  2. Why f1(v)+f2(v)f_1(v) + f_2(v) does NOT obey Maxwell’s law.

    The sum of two Maxwellians with different masses is not itself a Maxwellian — it cannot be written in the form Cv2e−αv2C v^2 e^{-\alpha v^2} with a single α\alpha. The combined distribution is a weighted mixture, not a pure Maxwell-Boltzmann shape. So option (A) is false.

  3. Why f1(v)f_1(v) and f2(v)f_2(v) are not the same. …

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