a) In a process on a system, the initial pressure and volume are equal to final pressure and volume, then

Explanation

when the gas is compressed in a short time, the process is adiabatic. Thus,
`T_(2)V_(2)^(gamma-1) = T_(1)V_(1)^(gamma-1)`
or, `T_(2) = T_(1)((V_1)/(V_2))^(gamma-1)`
`=(300K)xx[(800)/(200)]^(0.4) = 522K`
Rise in temperature = `T_(2) -T_(1) = 222 K`
(b) When the gas is compressed in a long time. the process is isothermal thus, the temperature remains equal to the temperature of the sourrounding that is `27^(@)C`. THe rise in temperature = 0.