One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is α^24R J mol…

Question

One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is α^24R J mol K^(-1); then the value of α will be _____ . (Assume that the given diatomic gas has no vibrational mode.)

MARKS Content Team · Accepted Answer

Correct answer: 3. Molecular specific heat at constant volume in terms of degrees of freedom is given by C_V=f/2R Molecular specific heat of mixture at constant volume is C_V/mix=(n_1C_V1+n_2C_V2)/(n_1+n_2) =(1×3R)/2+3×5R/21+3 =9R/4=α^24R ⇒α=3

One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is $\frac{α^{2}}{4} R J mol K^{- 1}$ ; then the value of $α$ will be _____ . (Assume that the given diatomic gas has no vibrational mode.)

Solution

Relevant Questions

One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is α24R J mol K-1; then the value of α will be _____ . (Assume that the given diatomic gas has no vibrational mode.)

Solution

Relevant Questions

One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is $\frac{α^{2}}{4} R J mol K^{- 1}$ ; then the value of $α$ will be _____ . (Assume that the given diatomic gas has no vibrational mode.)