Question

A monoatomic gas undergoes adiabatic process. Its volume and temperature are related as `TV^(P)` = constant. The value of p will be
`{:((1),1.33,(2),1.67),((3),0.67,(4),0.33):}`

Answer 1

For adiabatic process
`TV^(gamma-1)` = constant
for monoatomic gas `gamma` = 1.67
p = 1.67 - 1 = 0.67
`therefore "Ans"(3)`