One mole of a monoatomic gas is mixed with 3 moles of a heat of the mixture at constant volume ? Take `R=8.31 j mol e//K`.
For monoatomic gas, `C_(v)=3 R//2` and for diatommic gas, `C_(p)=5 R//2`.

Answer

`[18.7 j//mole//K]`

Explanation

`(c_(p))/(c_(v))=1.51, c_(p)=1.51 c_(v)`
Also, `c_(p)-c_(v)=(r)/(j)=(Pv)/(Tj)=(P(1//rho))/(Tj)`
`c_(p)-c_(v)=(1.013xx10^(6))/(1.234xx10^(-3)xx273xx4.2xx10^(7))`
`=0.0715`
`:. 1.51 c_(v)-c_(v)=0.0715`
or `0.51 c_(v)=0.0715`
or `c_(v)=(0.0715)/(0.51)=0.1404 cal. g^(-1) K^(-1)`,
`c_(p)=1.51 c_(v)=1.51xx0.1404`
`=0.2119 cal. g^(-1) K^(-1)`