Exams › JEE Main › Physics
A mixture of 2 moles of helium gas (atomic mass = 4u), and 1 mole of argon gas (atomic mass = 40u) is kept at 300 K in a container. The ratio of their rms speeds [V_rms (helium) / V_rms (argon)] is close to:
- 3.16
- 0.32
- 0.45
- 2.24
Correct answer: 3.16
Solution
The root mean square speed of a gas is inversely proportional to the square root of its molar mass. Since helium has a much lower molar mass than argon, its rms speed is significantly higher, resulting in a ratio of approximately 3.16.
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 →