Figure shows the adiabatic curve on log-log scale performed on a ideal gas. The gas must be:-

Options

A. Monoatomic
B. Diatomic
C. A mixture of monoatomic and diatomic
D. A mixture of diatomic and polyatomic

Answer

C

Explanation

For adiabatic process `TV^(gamma-1)` = constant
`implies log ` T + (gamma - 1) "log" V` = constant ` imlpies` slope = `-(gamma-1) = - ((5)/(10)) implies gamma = (3)/(2)`
For monoatomic gas `gamma =(5)/(3)`, " "` For diatomic gas `gamma = (7)/(5) " " "As" (7)/(5) lt gamma = (3)/(2) lt (5)/(3)`
Hence, the gas must be a mixture of monoatomic & diatomic gas.