Correct Answer - A
`M=4gm`
`V=22.4`litre
`C_(V)=5(J)/("Mole K")`
`V_("sound")=952m//sec`
`C_(P)=?`
`V_("sound")=sqrt((gammaPV)/(M))`
`gamma=(M)/(PV)V_("sound")^(2)=(C_(P))/(C_(V))`
`C_(P)=C_(V)((M)/(PV))V_("sound")^(2)`
`=5[(4xx10^(-3))/(10^(-5)xx22.4xx10^(-3))](952)^(2)`
`=(20)/(22.4)xx(952)^(2)xx10^(-5)`
`=809,200xx10^(-5)=8.09J//"moles k"`