Question

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in K)

Answer 1

The final temperature is 83 K.

Step-by-step explanation:

For an adiabatic process,

`T {V}^(\gamma - 1) = \text{constant}`

`\cfrac{{T}_(2)}{{T}_(1)} = {\left( \cfrac{{V}_(1)}{{V}_(2)} \right)}^(\gamma - 1)`

Given:-

`{T}_(1) = 275 \; K`

`{T}_(2) = T \left( \text{say} \right)`

`{V}_(1) = V`

`{V}_(2) = 6V`

`\gamma = \cfrac{5}{3} \;` (the gas is monoatomic)

`\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{(5)/(3) - 1}`

`\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{(2)/(3)}`

T = 275
`*` 0.30

T = 83 K.