The molar heat capacity C for an ideal gas going through a given process is given by `C=a/T`, where 'a' is a constant. If `gamma=C_p/C_v`, the work done by one mole of gas during heating from `T_0` to `eta T_0` through the given process will be

Options

A. `1/a ln eta`
B. `aln eta -((eta-1)/(gamma-1)) RT_(0)`
C. `aln eta-(gamma-1) RT_(0)`
D. none of these

Answer

2

Explanation

`Q=int_(T_(0))^(etaT_(0)) CdT=a ln (etaT_(0))/(T_(0))=aln eta`
`DeltaU=Cv DeltaT`
`=R/(r-1) (eta-1)T_(0)`
`omega=Q-DeltaU=alneta-[(eta-1)/(r-1)]RT_(0)`