From the relation `p=1/3 mnc_(rms)^2` show that the mean kinetic energy of a molecule is the same for all types of gases.

Find out the molecular kinetic energy of 1 mol of an ideal gas . Is it equal for all gases?

At a constant temperature, a vessel of 1 litre capacity contains 10^(23)N_(2) molecules. If the rms velocity of the molecules be 10^(3)m//s , then determine the total kinetic energy of the molecules and the temperature of the gas.

Use kinetic theory of gases to show that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

"The total kinetic energy of the molecules in an ideal gas with a volume V at pressure P and temperature T is equal to the total kinetic energy of the molecules present in the same volume of another ideal gas at the same pressure and at temperature 2T"-Justify the statement.

At 0^(@)C the kinetic energy of 1O_(2) molecule is 5.62xx10^(-14) erg. Determine avogadro's number.

According to the kinetic theory of gases, the average kinetic energies of O_(2) and N_(2) molecules are the same at a particular temperature. State whether the rms velocities of the molecules of the two gases at a given temperature will be the same or not.

Prove that c_(rms)=sqrt((2E)/(M)) [E=total kinetic energy of the molecules of 1 mol of a gas, M=molar mass of the gas, c_(rms)= root mean square velocity of gas molecule].

If, at a given temperature, the totla kinetic energy of the molecules in unit volume of an ideal gas be E, show that the pressure of the gas, P=2//3E .

At a given temperature, the average kinetic energy of H_(2) molecules is 3.742 kJ*mol^(-1) . Calculate root mean square velocity of H_(2) molecule at this temperature.

Write down the expansion for pressure of an ideal gas according to kinetic theory of gases. Given that at absolute temperature T, the average kinetic energy of a molecule of an ideal gas is 3/2(R/N)T , where R is the universal gas constant and N is the Avogadro number Determine the equation for ideal gases from his.