Exams › JEE Main › Physics
Number of molecules in a volume of 4 cm³ of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to ? (Given, mean kinetic energy of a molecule (at T) is 4 × 10⁻¹⁴ erg, g = 980 cm/s², density of mercury = 13.6 g/cm³)
- 5.8 × 10¹⁸
- 5.8 × 10¹⁶
- 4.0 × 10¹⁸
- 4.0 × 10¹⁶
Correct answer: 4.0 × 10¹⁸
Solution
P = rho*g*h = 13.6*980*2 = 26656 dyne/cm^2; PV = 26656*4 = 1.066e5 erg. kT = (2/3)(4e-14) = 2.67e-14 erg. N = PV/kT = 1.066e5/2.67e-14 ~ 4.0e18.
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 →