A gas is suddenly compressed to `1/4th` of its original volume. Caculate the rise in temperature when original temperature is `27^(@)C. gamma= 1.5`.

Options

A. 222.33 K
B. 233.33 K
C. 244.33 K
D. 255.33 K

Answer

A

Explanation

Here `T_(1)=27^(@)C`=27+273 =300K
`V_(2)=(V_(1))/(4)`
As the gas is compressed suddenly then the process is adiabatic then
`TV^(gamma-1)`=constant `rArr T_(1)V_(1)^(gamma-1)=T_(2)V_(2)^(gamma-1)`
or
Hence change in temperature
`rArrT =T_(2)-T_(1)=522.33-300=222.33 K`