A diatomic gas initally at `18^(@)C` is compressed adiabtically to one- eighth of its original volume. The temperature after compression will be

Options

A. `18^(@)` C
B. `887^(@)` C
C. `327^(@)` C
D. `395.5^(@)` C

Answer

D

Explanation

`T_(1) = 18^(@) C = (273 + 18) = 291 K`
and `V_(2) =V_(1)//18`
We know that `TV^(gamma-1)` = constant
or `T_(2)V_(2)^(gamma-1)=T_(1)^(gamma-1)`
`therefore T_(2) =(V_(1)/V_(2))^(gamma -1) = 291 xx (8)^(1.4-1) = 668.5 K = 395.5^(@)` C