A mixture of 1 mole helium and 1 mole nitrogen is enclosed in a vessel of constant volume at 300 K. Find the quantity of heat absorbed by the mixture if the root mean square speed of its molecules get doubled. Give your answer in terms of universal gas constant R.

A mass M = 15 g of nitrogen is enclosed in a vessel at temperature T = 300 K. What amount of heat has to be transferred to the gas to increase the root-mean-square velocity of molecules 2 times ?

About 0.014 kg nitrogen is enclosed in a vessel at temperature of 27^(@)C How much heat has to be transferred to the gas to double the rms speed of its molecules ? (R = 2 cal//mol K)

A mixture of two gases is contained in a vessel. The Gas 1 is monoatomic and gas 2 is diatomic and the ratio of root mean square speeds of the molecules of two gases is

A fixed mass of gas at constant pressure occupies a volume V. The gas undergoes a rise in temperature so that the root mean square velocity of its molecules is doubled. The new volume will be

14 g of nitrogen is contained in a vessel at 300 K. How much heat should be taken out of the gas to half the rms speed cf its molecules ? R=2 cal/mol K:

Two moles of helium are mixed with n moles of hydrogen. The root mean spure (rms) speed of the gas molecules in the mexture is sqrt2 times the speed of sound in the mixture. Then value of n is

An ideal diatomic gas occupies a volume V_1 at a pressure P_1 The gas undergoes a process in which the pressure is proportional to the volume . At the end of process the root mean square speed of the gas molecules has doubled From its initial value then the heat supplied to the gas in the given process is

Four moles of hydrogen, 2 moles of helium and 1 mole of water form an ideal gas mixture. What is the molar specific heat at constant pressure of mixture ?

A vessel of volume V contains n_(1) moles of oxygen and n_(2) moles of carbon dioxide at temperature T . Assuming the gases to be ideal, find, (a) the pressure of the mixture, (b) the mean molar mass of the mixture.