Exams › JEE Main › Physics
The value closed to the thermal velocity of a Helium atom at room temperature (300K) in ms⁻¹ is -
[kB = 1.4 × 10⁻²³ J/K; mHe = 7 × 10⁻²⁷ kg]
[JEE-Main On line-2018]
- 1.3 × 10⁴
- 1.3 × 10⁵
- 1.3 × 10²
- 1.3 × 10³
Correct answer: 1.3 × 10³
Solution
The thermal velocity of a gas particle can be calculated using the formula v = sqrt(3kT/m), where k is the Boltzmann constant, T is the temperature, and m is the mass of the particle. Substituting the values for helium at room temperature yields a result close to 1.3 × 10³ ms⁻¹, making this the correct option.
Related JEE Main Physics questions
- An ideal gas is enclosed in a thermally insulated sealed container. During an adiabatic expansion, the mean interval between successive molecular collisions varies with the volume V as V^q. What is the value of q in terms of γ = Cp/Cv?
- At 300 K, the root mean square speed of hydrogen molecules is 1930 m/s. What will be the r.m.s. speed of oxygen molecules at 1200 K?
- A gaseous mixture contains 2 moles of oxygen and 4 moles of argon at temperature T. If vibrational degrees of freedom are ignored, what is the total internal energy of the mixture?
- If all intermolecular attractions were removed, what volume would be occupied by the molecules present in 4.5 g of water at standard temperature and pressure?
- An ideal gas sample has volume V, pressure P, and absolute temperature T. If the mass of one molecule is m, which expression gives the density of the gas?
- A vessel at temperature T contains N molecules of gas A, each having mass m, and 2N molecules of gas B, each having mass 2m. If the mean square speed of molecules of gas B is V2 and that of gas A is V1, then the ratio V1/V2 equals
⚔️ Practice JEE Main Physics free + battle 1v1 →