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 adiabatic procvess then
`T_(1)V_(1)^(gamma-1)=T_(2)v_(2)^(gamma-1)`
Putting `T_(1)=273+20=293K, V_(2)=8V_(1)`
`(293)(V_(1))^(gamma-1)=T_(2)(8V_(1))^(gamma-1)`
`293=T_(2)^(gamma-1)`
`T_(2)=(293)/(8^(gamma-1))=(293)/(8^(5)/(3)-1))`
`T_(2)=(293)/(2^(3))^(2//3)=(293)/(4) =73.25 K`