Q.A sample of ideal gas is at equilibrium. Which of the following quantity is zero?

Definition and formula. The mean (or average) speed, $\bar v$, is the plain arithmetic average of every individual molecule's speed in the gas -- unlike $v_{rms}$, it averages the speeds directly rather than averaging their squares first. Derived from the Maxwell speed distribution function, its formula is

Where it sits among the three speeds. Because $1.60<1.73$, $\bar v$ is always smaller than $v_{rms}$ for the same gas at the same temperature -- specifically, $\bar v\approx0.92\,v_{rms}$. This is a genuinely important distinction: many students instinctively confuse 'root mean square speed' with a simple 'average speed', but the two are conceptually different averaging procedures over the same underlying speed distribution, and they give numerically different (though closely related) answers.

A vector subtlety worth noting. The average speed $\bar v$ of a gas at equilibrium is a nonzero positive number -- molecules genuinely are moving, on average, at hundreds of metres per second. But the average velocity (a vector, not a scalar) of the same gas at equilibrium is exactly zero, because the postulates of kinetic theory require completely random motion with no preferred direction, so every molecule's velocity is, on average, exactly cancelled by some other molecule moving the opposite way. Confusing 'speed' (a scalar magnitude, always positive) with 'velocity' (a vector, which can average to zero even when nothing is actually motionless) is one of the most common conceptual traps in this topic.