Two ideal gases, one monoatomic and another diatomic, are having same pressure P, volume V and temperature T. Select the correct optionThe translational kinetic energy for monatomic gas is 32PV but for diatomic gas it is 52PV.The translational kinetic energy for both monatomic and diatomic gas will be same.Average velocity of each molecule of any type of gas will be proportional to T5−−−√, where T is absolute temperature.Average energy for one molecule per degree of freedom for monatomic gas is 12kT but that for diatomic it is 2×12kT.

The average translational kinetic energy per molecule of an ideal gas at 0ºC (k =1.38 × 10–23 J/K)  is –

At  31 °C, the average translational kinetic energy of a gas molecule is K. If the temperature is now increased to 59 °C, the average translational kinetic energy of a molecule, in terms of K,  will be closest to

Assertion (A) : The ratio of specific heat of a gas at constant pressure and specific heat at constant volume for a diatomic gas is more than that for a monatomic gas.Reason (R) : The molecules of a monatomic gas have more degree of freedom than those of a diatomic gas.

What is the average kinetic energy per particle in 1.00 mole of an ideal gas at a pressure of 1.50 atmospheres and a volume of 2.00 liters?

A quantity of a monatomic ideal gas undergoes a process in which both its pressure and volume are increased by a factor of n = 3 as shown in the figure below. What is the energy absorbed by heat into the gas during this process? Hint: The internal energy of a monatomic ideal gas at pressure P and occupying volume V is given by U = 32PV. (Give your answer as a multiple of P0V0.) Your response differs from the correct answer by more than 10%. Double check your calculations. ✕ P0V0