At what temperature will the average velocity of oxygen molecules be sufficient to escape from the earth. Given mass of oxygen molecule `= 5.34 xx 10^(-26) kg`. Boltzmann constant, `k = 1.38 xx 10^(-23) J "molecule"^(-1) K^(-1)`. Escape velocity of earth `= 11.0 km s^(-1)`.

At what temperature the kinetic energy of a molecule will be equal to 2.8 xx 10^(-20) J ? Boltzmann constant (k_(B)) = 1.4 xx 10^(-23) J "molecule"^(-1)K^(-1)

At what temperature , will the rms speed of oxygen molecules be sufficient for escaping from the earth ? Taken m = 2.76 xx 10^(-26) kg, k = 1.38 xx 10^(-23) J//K and v_(e) = 11.2 km//s .

At what temperature the root mean square velocity is equal to escape velocity from the surface of earth for hydrogen and for oxygen ? Given radius of earth = 6.4 xx 10^(6) m , g = 9.8 ms^(-2) . Boltzmann constant = 1.38 xx 10^(-23) J "molecule"^(-1) .

The diameter of a gas molecule is 2.4 xx 10^(-10) m . Calculate the mean free path at NTP. Given Boltzmann constant k = 1.38 xx 10^(-23) J molecule^(-1) K^(-1) .

At what temperature, hydrogen molecules will escape from the earth's surface ? (Take, radius of earth R_(e)=6.4xx10^(6)m , mass of hydrogen molecule m=0.34 xx 10^(-26) kg, Boltzmann constant k=1.38xx10^(-23)JK^(-1) and acceleration due to gravity = 9.8 xx ms^(-2) ) also take rms speed of gas as v_(rms) = sqrt((3kT)/(m)) .

Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at 0^(@)C. Mass of a helium molecule =6.64xx10^(-27)kg and Blotzmann constant =1.38xx10^(-23) J K^(-1).

Calculate the temperature at which the average K.E. of a molecule of a gas will be the same as that of an electron accelerated through 1 volt. Boltzmann constant = 1.4 xx 10^(-23) J "molecule"^(-1) K^(-1) , charge of an electron = 1.6 xx 10^(-19)C .

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to: [Boltzmann constant k _ B = 1.38 xx 10 ^( -23) J//K Avogadro Number N _ A = 6.02 xx 10 ^( 26) // Kg Radius of Earth : 6.4 xx 10 ^( 6 ) m Gravitational acceleration on Earth = 10 ms ^( -2) ]