An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity C remains constant. It during this process the relation the pressure p and volume V is given by pV^n=constant, then n is given by (here C_p and C_v are molar specific heat at constant pressure and constant volume respectively).

An ideal gas undergoes a quasistatic, reversible process in which its

1.	An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity C remains constant. It during this process the relation the pressure p and volume V is given by pV^n=constant, then n is given by (here C_p and C_v are molar specific heat at constant pressure and constant volume respectively).
Answer» `n=C_p/C_v` `n=(C-C_p)/(C-C_v)` `n=(C_p-C)/(C-C_v)` `n=(C-C_v)/(C-C_p)` Solution :Here `PV^n=k` (constant)…….(1) For 1 mol of ideal gas `pV=RT`……..(2) Dividing (1) by (2) we get `V^(n-1) T=k/R` `THEREFORE((dV)/(dT))=V/((n-1)T)=V/((1-n)T)` According to first law of THERMODYNAMICS `dQ=C_vdT+pdV` `therefore(dQ)/(dT)=C_V+p((dV)/(dT))=C_v+(pV)/((1-n)T)=C_v+R/(1-n)` Hence thermal capacity, `C=C_v+R/(1-n)` or,`1-n=R/(C-C_v)` or,`n=1-R/(C-C_v)=(C-(C_v-R))/(C-V_v)=(C-C_p)/(C-C_v)`[`because C_p-C_v=R`]