The kinetic energy of a molecule of oxygen at `0^(@)C` is `5.64 xx 10^(-21) J`. Calculate Avogadro's number. Given `R = 8.31 J "mol^(-1)K^(-1)`.

The kinetic energy of a molecule of hydrogen at 0^(@) is 5.64 xx 10^(-21) J . Calculate Avogadro's number. Take R = 8.31 xx J "mole"^(-1) K^(-1)

Kinetic energy of oxygen molecule at 0^(@) C is 9.4 xx 10^(-21) j . Calculate Avogadro's number, when R = 8.31 J "mole"^(-1) K^(-1) .

The kinetic energy of two moles of N_(2) at 27^(@) C is (R = 8.314 J K^(-1) mol^(-1))

The average kinetic energy of a gas molecule at 0^(@)C is 5.621xx10^(-21) J . Calculate Boltzmann costant. Also calculate the number of molecules present in one mole of the gas.

Calculate kinetic energy of a gram molecule of oxygen at 127^(@)C . Value of Boltzmann constant = 1.381 xx 10^(-23) J K^(-1) . Avogardro's number = 6.022 xx 10^(23) per gm-mole.

The average kinetic energy of gas molecule at 27^(@)C is 6.21xx10^(-21) J. Its average kinetic energy at 127^(@)C will be

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)

The kinetic energy of 1 kg of oxygen at 300 K is 1.169 xx 10^(5) J . Find the kinetic energy of 2 kg of oxygen at 500 K.

The average kinetic energy of a hydrogen molecule at 27^(@)C is 9.3 xx 10^(-21) J . The mass of hydrogen molecule is 3.1 xx 10^(-27) kg . (i) Determine the average kinetic en-ergy at 227^(@)C . (ii) Determine the root mean square speed of hydrogen molecule at 27^(@)C .