Correct Answer - A
Here, `C_(upsilon)=5.0JK^(-1)mol^(-1), upsilon=952ms^(-1)`
As `4.0g` of gas occupies `22.4` litres at `N.T.P.,` hence molar mass of gas, `m=4g//mol`
Velocity of sound,
`upsilon=sqrt((gamma P)/(rho))=sqrt((gammaP)/(m//V))=sqrt((gammaPV)/(m))=sqrt((gammaRT)/(m))` ,brgt or `upsilon^(2)=(gammaRT)/(m)`
or `gamma=(upsilon^(2)m)/(RT)=((952)^(2)xx(4xx10^(-3)))/(8.3xx273)=1.6`
`gamma=(C_(p))/(C_(v))` or `C_(p)=gammaC_(v)=1.6xx5`
`=8.0JK^(-1)mol^(-1)`