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One mole of a monatomic gas is mixed with 3 moles of a diatomic gas. What is the molar specific heat of the mixture at constant volume?

Answer»

Solution :For monoatomic gas , `C_(v)=(3)/(2)R`, `N=1` mole
For diatomic gas, `C_(v)'=(5)/(2)R`, `n'=3` mole
From conservation of energy , the molecular SPECIFIC heat of the mixture is
`C_(v)'=(n(C_(v))+n'(C_(v)'))/((n+n'))`
`=(1XX(3)/(2)R+3xx(5)/(2)R)/((1+3))=(9)/(4)R`
or `C'v=(9)/(4)xx8.31=18.7 J mol^(-1)K^(-1)`


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