11.
The number of molecules dN with speeds lying between v and v+dv of a classical gas is
given by dN = ρ(v) dv. The distribution ρ(v) is shown for two systems in the figure. You
are told that ρ(v) for system 2 is twice that of system 1 for all v. What can we say about
the two systems?
a. System 1 is twice as hot as System 2
b. The molecules of System 2 are twice as heavy as those of System 1
c. The molecules in System 2 are moving twice as fast as those in System 1
d. System 2 has twice the number of molecules as System 1
Introduction
In kinetic theory, the Maxwell–Boltzmann speed distribution of a classical gas describes how many molecules have speeds between v and v+dv at a given temperature. When two gases share the same distribution shape but one curve is uniformly scaled by a constant factor, this directly reflects a change in the total number of molecules rather than a change in temperature or molecular mass.
The question states that the number of molecules with speeds between v and v+dv is given by dN = ρ(v) dv, and that for System 2, ρ2(v) = 2ρ1(v) for all v. This means every speed interval contains exactly twice as many molecules in System 2 as in System 1.
Key Idea: Normalization of ρ(v)
The total number of molecules N in a system is obtained by integrating the distribution over all speeds:
N = ∫0∞ ρ(v) dv.
If ρ2(v) = 2ρ1(v) for all v, then
N2 = ∫0∞ρ2(v) dv = ∫0∞2ρ1(v) dv = 2N1, so System 2 contains twice as many molecules as System 1, while the functional dependence on v (and therefore the shape, temperature dependence, and mass dependence) remains unchanged.
Option (d): Correct Statement
For every small speed interval [v, v+dv], System 2 has dN2 = ρ2(v) dv = 2ρ1(v) dv = 2 dN1 molecules. Since this factor of 2 is the same for all speeds, integrating over all speeds gives N2 = 2N1, which matches option (d). This is the only statement justified directly by ρ2(v) = 2ρ1(v) for all v.
Therefore, option (d) is correct: System 2 has twice the number of molecules as System 1, while both systems have the same temperature and the same molecular mass distribution.
Option (a): Temperature Comparison
For an ideal classical gas, the Maxwell–Boltzmann speed distribution has the form ρ(v) ∝ v² exp(−mv² / 2kBT), where m is particle mass and T is the absolute temperature. Changing the temperature changes the shape of the distribution: higher T shifts the peak to higher speeds, lowers the maximum value, and broadens the curve.
In the given figure, the curves for System 1 and System 2 have the same shape and peak position in v, with System 2 simply having a higher amplitude at every speed. Because the shape and most probable speed are unchanged, the temperature must be the same for both systems, so option (a) is false.
Option (b): Molecular Mass Comparison
The most probable speed vmp for a Maxwell–Boltzmann speed distribution is vmp = √(2kBT / m), which depends on both temperature and molecular mass. If the mass increased while temperature stayed fixed, the peak of the speed distribution would shift to lower speeds and the curve shape in terms of v would change.
In the problem, the speed distribution curves for both systems peak at the same speed and have the same width, indicating that m and T are identical and only the overall magnitude changes. Hence, the molecules in System 2 are not heavier; they are the same species as in System 1, so option (b) is false.
Option (c): Molecular Speed Comparison
If molecules were moving twice as fast on average, the peak and characteristic speeds of the Maxwell–Boltzmann distribution would shift to larger v. The figure, however, shows that both systems share the same distribution along the velocity axis; only the vertical scale differs.
That means all characteristic speeds (most probable, mean, and rms) are the same in both systems. Therefore, molecules in System 2 are not moving twice as fast; they have the same speed distribution as those in System 1, so option (c) is false.
Summary Table of Options
| Option | Statement | Correct? | Reason (brief) |
|---|---|---|---|
| a | System 1 is twice as hot as System 2 | No | Same distribution shape and most probable speed imply same temperature; temperature changes would distort the curve shape. |
| b | Molecules of System 2 are twice as heavy as those of System 1 | No | Changing mass at fixed temperature shifts peak speed; identical peak positions show molecular mass is unchanged. |
| c | Molecules in System 2 move twice as fast as those in System 1 | No | If speeds doubled, peak and average speeds would move to higher v; here only the vertical scale changes. |
| d | System 2 has twice the number of molecules as System 1 | Yes | ρ2(v) = 2ρ1(v) for all v means every speed interval has twice as many molecules, so total N2 = 2N1. |
