Calculate the ratio of specific heats for nitrogen. Given that the specific heat of nitrogen at cinstant pressure `=0.236` cal `g^(-1) K^(-1)` and density at S.T.P. is `0.001234 g//c c`. Atmospheric pressure`=1.01xx10^(6) dyn e//cm^(2)` .

Answer

`[1.434]`

Explanation

For `1` gram of nitrogen
`c_(p)-c_(v)=(r)/(j)=(Pv)/(Tj)=(P)/(rho Tj)" " [thereforev=(mass)/(density)=(1)/(rho)]`
`c_(v)=c_(p)-(P)/(rho Tj)`
`=0.236-(1.01xx10^(6))/(0.001234xx273xx(4.2xx10^(7))`
`=0.236-0.0714=0.1646 cal g^(-1) K^(-1)`
`(c_(p))/(c_(v))=(0.236)/(0.1646)=1.434`