Question

Calculate the fall in temperature of helium initially at `15^(@)C`, when it is suddenly expanded to `8 times` its original volume `(gamma=5//3)`.

Answer 1

Here, `T_(1)=273+15=288 K` ,
`T_(2)=?, V_(2)=8V_(1), gamma=5//3`
As expansion is sudden //adiabatic
`:. T_(2)V_(2)^(gamma-1)= T_(1)V_(1)^(gamma-1)`
`:. T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1) = 288((V_(1))/(8V_(1)))^((5/3-1))`
or `logT_(2)= log288+2/3log(1/8)`
`log288+2/3[log1-log8]`
`logT_(2)= 2.4594+2/3[0-0.90031]`
`1.8573`
`T_(2)= antilog 1.8573= 71.99K`
`:.` fall in temperature of helium
`T_(1)-T_(2)=288-71.99= 216.01K`
`=216.01^(@)C`
`( :. "in magnitude", 1^(@)C=1K)`