The fall in temperature of helium gas initially at `20^(@)` when it is suddenly expanded to 8 times its original volume is `(gamma=(5)/(3))`

Options

A. `70.25 K `
B. `71.25 K`
C. `72.25 K `
D. `73.25 K `

Answer

D

Explanation

Since gas is suddenly expanded it means the process is adiablatic process, then
`T _(1) V _(1) ^( gamma -1) = T _(2) V _(2) ^( gamma -1)`
Putting
`T _(1) = 273 + 20 = 293 K, V _(2) = 8V _(1)`
`(293) (V _(1)) ^( gamma -1) = T _(2) ( 8 V _(1)) ^( gamma -1)`
`293 = T _(2) 8 ^( gamma -1)`
`T _(2) = ( 293)/( 8 ^( gamma -1))`
`= ( 293)/( 8 (5)/(3)-1)" "[therefore gamma = (5)/(3)]`
`T _(2) = ( 293)/( 8 ^( 2//3)) = ( 293)/((2^(2)) ^( 2//3)) = ( 293)/(4) = 73.25 K`