When 1g `H_(2) ` is mixed with 1 g He than find
(I ) ` gamma (=( C_(p))/( c_( v )))` of the mixture
` (ii) c_(p) ` ( molar specific heat at constant pressure ) of the gas mixture

Explanation

`1g H_(2) = (1)/(2) mol H_(2) , 1 g He = 1/4 mol He `
For 1 mol of hydrogen which is diatomic gas ,
` C_(v ) . =(5)/(2) R and C_(p) . =5/2 R + R=7/2 R`
For 1 mol of helium which is monoatomic gas
`C_(v) . . =(3)/(2) R and C_(p ) . . =3/2 R+R =5/2 R `
`therefore ` for the mixture ,
`C_(v ) = ( n_(1)C_v . +n_(2)C_(v ). .)/(n_(1) +n_2)=(1/2xx 5/2 R+1/4xx 3/2 R)/(1/2 +1/4)= (13)/(6) R `
Similarly , `C_(p) = ( n_(1 ) C_(p) . +n_(2) C_(p). .)/( n_(1)+n_(2))=(1/2 xx7/2 R 1/4 xx5/2R)/( 1/2 +1/4)=(19)/(6)R`
Hence (i ) ` gamma =(c_p )/(c_(v))=(19)/(13)(ii ) C_(p) =(19)/(6) R `