The degree of freedom per molecule for a gas is `6`. At constant pressure work done by gas is `25J`. The heat supplied to the gas for this purpose is `25x` joule them `x` is:

Explanation

`lambda=1+(2)/(f)=1+(1)/(3)`
`lambda=(4)/(3)`
As we know `DeltaW=DeltaQ-DeltaU`
`therefore (DeltaW)/(DeltaQ)=(DeltaQ-DeltaU)/(DeltaQ)=1-(C_(v))/(C_(p))`
`therefore (DeltaW)/(DeltaQ)=1-(1)/(lambda)=(1)/(4)`
`therefore DeltaQ=4.DeltaW`
`DeltaQ=100J.