Question

An ideal monatomic gas at 300 k expands adiabatically and reversibly to twice its volume. what is its final temperature?

Answer 1

In an adiabatic process, the following relationship holds:

`TV^(\gamma -1) = cost.`
where T is the gas temperature, V is the volume and
`\gamma` is the adiabatic index, which is equal to
`\gamma = (5)/(3)` for a monoatomic gas.

We can re-write the equation as

`T_1 V_1^(\gamma -1) = T_2 V_2^(\gamma -1)`
where the labels 1,2 refer to the initial and final conditions of the gas.
Let's rewrite it for
`T_2`, the final temperature:

`T_2 = T_1 ( (V_1)/(V_2) )^(\gamma-1)`

We can now substitute the initial temperature, T1=300 K, and
`V_2 = 2V_1`, because the final volume is twice the initial one. So we find the value of the final temperature:

`T_2 = 300 K( (1)/(2))^{ (2)/(3) } =189 K`