An ideal monoatomic gas at 300K expands adiabatically to twice its volume. What is the final temperature?

Options

A. `300 sqrt(2)`
B. `300 sqrt(3)`
C. `300 ((1)/(2))^(2//3)`
D. `300 (2)^(2//3)`

Answer

C

Explanation

For an ideal monoatomic gas, the heat capacity `T_(i) T_(f)` , =initial and final temperatures of gas
ratio `gamma = (5)/(3)`
In adiabatic process
`TV^(-1) ` = constant
`T_(f) V_(f)^(-1) = T_(i) V_(i)^(gamma-1) implies T_(f) = T_(i) (V_(i)/(V_(f)))^(gamma-1 ) =300 ((1)/(2))^((5)/(3)-1)`
`=300 ((1)/(2))^(2//3)`