The ratio of slope of adiabatic to that of an isothermal process is:

Options

A. `(1)/(gamma)`
B. `gamma`
C. `gamma-1`
D. None of the above.

Answer

B

Explanation

For isothermal process PV=constant Differentiating both sides
PdV + VdP =0
`therefore (dP)/(dV) =-(P)/(V)`
Thus, slope of isothermal process `((dP)/(dV))=-PV" ….(i)"`
For adiabatic process `PV^(gamma)=` constant differentiating both sides
`P. gamma. V^(gamma-1)+V^(gamma). dP=0`
`(dP)/(dV)=(gamma P)/(V)`
`therefore` slope of adiabatic process `((dP)/(dV))=(-gamma P)/(V)`
Dividing eqn. (ii) and (i)
`("Slope of adiabatic process")/("Slope of isothermal process")=gamma`
Hence, correct choice is (b).