Question

At temperature \( T \), the probability of ideal gas molecules with mass \( m \) having speed \( v \) is given by \( P(v) \). Two quantities \( I_1 \) and \( I_2 \) are defined in terms of \( P(v) \) as follows:
\[ I_1 = \int_0^\infty P(v) v^3 \, dv \quad \text{and} \quad I_2 = \int_0^\infty P(v) v \, dv \] Which of the following relationship is true?

• A. \( I_1 = I_2 \)
• B. \( I_1 = \frac{I_2}{2} \)
• C. \( I_1>I_2 \)
• D. \( I_1<I_2 \)

Hint:

When dealing with integrals involving probability distributions, moments with higher powers of the variable typically decay faster, which means they tend to be smaller than moments with lower powers.

Answer 1

The Correct Option is D

The quantities \( I_1 \) and \( I_2 \) are integrals involving the speed distribution function \( P(v) \), and they represent different moments of the speed distribution. From the Maxwell-Boltzmann distribution, we know that the speed probability distribution \( P(v) \) is proportional to \( v^2 e^{-\frac{mv^2}{2kT}} \), where \( m \) is the mass of the molecules and \( T \) is the temperature. ### Step 1: Evaluate the relationship between \( I_1 \) and \( I_2 \). - \( I_1 = \int_0^\infty P(v) v^3 \, dv \) is the moment that involves the \( v^3 \) term, which has a higher dependence on the velocity and is affected more by higher speeds. - \( I_2 = \int_0^\infty P(v) v \, dv \) involves the \( v \) term and is less dependent on higher velocities. In general, for any given distribution, the moment involving \( v^3 \) (which represents higher-order moments) will be smaller than the moment involving \( v \) since the integrand in \( I_1 \) decays faster than that in \( I_2 \) due to the higher powers of \( v \). This means that: \[ I_1<I_2 \] Thus, the correct relationship is \( I_1<I_2 \).