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A gas of molecules each having mass' m' is in thermal equilibrium at a temperature T. Let v, and v, be the Cartesian components of velocity v of a molecule. The mean value of (vx+ vy³) is (a) 0 (b) kT (c) 2kgT (a) kT?
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A gas of molecules each having mass' m' is in thermal equilibrium at a...
Thermal Equilibrium and Mean Value
In thermal equilibrium, the gas molecules have a distribution of velocities described by the Maxwell-Boltzmann distribution. The mean value of a quantity in this distribution can be calculated by integrating the quantity multiplied by the probability density function over all possible velocities.

Calculating the Mean Value of (vx vy³)
To find the mean value of (vx vy³), we need to calculate the integral of (vx vy³) multiplied by the probability density function f(vx, vy) over all possible velocities.

The probability density function for the velocity of a molecule in the x and y directions is given by:
f(vx, vy) = (m/(2πkT)) * exp((-m(vx²+vy²))/(2kT))

Integrating (vx vy³) * f(vx, vy) over all possible velocities gives:
∫∫(vx vy³) * f(vx, vy) d(vx) d(vy)

To simplify the integral, we can use the fact that the Maxwell-Boltzmann distribution is isotropic, meaning it does not depend on the direction of the velocity.

Therefore, the probability density function can be written as:
f(v) = (m/(2πkT)) * exp(-mv²/(2kT))

Using this simplified form, we can rewrite the integral as:
∫∫(vx vy³) * f(v) d(vx) d(vy)

Since the integral is over all possible velocities, we can change the variables to polar coordinates:
∫∫(vx vy³) * f(v) d(vx) d(vy) = ∫∫(vr cosθ vr sin³θ) * f(v) r dθ dr

Integrating over θ from 0 to 2π gives:
∫∫(vr cosθ vr sin³θ) * f(v) r dθ dr = ∫(vr 0) ∫(2π 0)(vr cosθ vr sin³θ) * f(v) dθ dr

Simplifying the integrals gives:
∫(vr 0) ∫(2π 0)(vr cosθ vr sin³θ) * f(v) dθ dr = ∫(vr 0) 2πvr * ∫(2π 0)(cosθ sin³θ) * f(v) dθ dr

The inner integral over θ is zero for both terms, since the integrands are odd functions and we are integrating over the range (-π,π).

Therefore, the mean value of (vx vy³) is zero:
∫(vr 0) 2πvr * ∫(2π 0)(cosθ sin³θ) * f(v) dθ dr = ∫(vr 0) 2πvr * (0) dr = 0

Conclusion
The mean value of (vx vy³) for a gas of molecules in thermal equilibrium at temperature T is zero. This result is derived from the isotropy of the Maxwell-Boltzmann distribution and the properties of the integrals involved in calculating the mean value.
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A gas of molecules each having mass' m' is in thermal equilibrium at a temperature T. Let v, and v, be the Cartesian components of velocity v of a molecule. The mean value of (vx+ vy³) is (a) 0 (b) kT (c) 2kgT (a) kT?
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