Question

one mole of a monoatomic gas is mixed with three moles of a diatomic gas. if the molecular specific heat of the mixture at constant volume is then the value of will be: (assume that the given diatomic gas has no vibrational mode.)

Answer 1

The molecular specific heat at constant volume for the mixture of one mole of monoatomic gas and three moles of diatomic gas is (9/8)R.

Step-by-step explanation:

To determine the molar specific heat at constant volume, Cv, for a mixture of one mole of monatomic gas and three moles of diatomic gas without vibrational modes, we use the degrees of freedom for each gas type. A monatomic gas has 3 degrees of freedom (translational), and a diatomic gas, assuming it has no vibrational modes, has 5 degrees of freedom (3 translational and 2 rotational). Using the formula Cv = (d/2)R, where d is the number of degrees of freedom and R is the universal gas constant, we calculate the molar specific heat for each gas. The average molar specific heat of the mixture, Cvmix, is then found by taking the mole-fraction weighted average of the individual specific heats.

For the monatomic gas with 3 degrees of freedom, Cv = (3/2)R. For the diatomic gas with 5 degrees of freedom, Cv = (5/2)R. Since there are three times as many moles of diatomic gas as monatomic gas in the mixture, their contributions to the average specific heat are weighted accordingly. Thus, the specific heat of the mixture is:

Cvmix = [(1*(3/2)R) + (3*(5/2)R)] / 4

Simplifying, we get:

Cvmix = [(3/2R) + (15/2R)] / 4 = (18/2R) / 4 = (9/2R) / 4 = (9/8)R

Therefore, the molecular specific heat of the mixture at constant volume is (9/8)R.