Home
Class 11
PHYSICS
Temperature of diatomic gas is 300 K. If...

Temperature of diatomic gas is `300 K`. If moment of intertia of its molecules is `8.28 xx 10^-38 g- cm^2`. Calculate their root mean square angular velocity.

Promotional Banner

Similar Questions

Explore conceptually related problems

As adiabatic vessel contains n_(1) = 3 mole of diatomic gas. Moment of inertia of each molecule is I = 2.76 xx 10^(-46) kg m^(2) and root-mean-square angular velocity is omega_(0) = 5 xx 10^(12) rad//s . Another adiabatic vessel contains n_(2) = 5 mole of a monatomic gas at a temperature 470 K . Assume gases to be ideal, calculate root-mean-square angular velocity of diatomic molecules when the two vessels are connected by a thin tube of negligible volume. Boltzmann constant k = 1.38 xx 10^(-23) J//:"molecule" .

As adiabatic vessel contains n_(1) = 3 mole of diatomic gas. Moment of inertia of each molecule is I = 2.76 xx 10^(-46) kg m^(2) and root-mean-square angular velocity is omega_(0) = 5 xx 10^(12) rad//s . Another adiabatic vessel contains n_(2) = 5 mole of a monatomic gas at a temperature 470 K . Assume gases to be ideal, calculate root-mean-square angular velocity of diatomic molecules when the two vessels are connected by a thin tube of negligible volume. Boltzmann constant k = 1.38 xx 10^(-23) J//:"molecule" .

The temperature of a gas consisting of rigid diatomic molecules is T = 300 k . Calculate the angular root mean square velocity of a rotating molecule if its moment of inertia is equal to I = 2.1.10^-39 g.cm^2 .

The temperature of a gas consisting of rigid diatomic moleculoes is T = 300 K. Calculate the angular root-mean square velocity of a rotating molecules if its moment of inertia is I = 2.0 xx 10^(-40) kg m^(2) .

The temperature of a gas consisting of rigid diatomic moleculoes is T = 300 K. Calculate the angular root-mean square velocity of a rotating molecules if its moment of inertia is I = 2.0 xx 10^(-40) kg m^(2) .