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.

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)

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

 

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

T  =  275 $\times$ 0.30

T  =  83 K.