Exams › JEE Main › Physics
A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is:
- 11
- 15
- 20
- 13
Correct answer: 15
Solution
The total internal energy of an ideal gas mixture can be calculated using the formula U = (f/2) * nRT, where f is the degrees of freedom and n is the total number of moles. For oxygen, which has a rigid bond, the degrees of freedom is 5 (3 translational and 2 rotational), and for argon, it is 3 (only translational). Thus, the total internal energy for the mixture is calculated as (5/2 * 3 + 3/2 * 5)RT = 15RT.
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 →