A perfect gas in a cylinder provided with frictionless piston is at `37^(@)C`. The gas is cooled at constant pressure so that its volume becomes `(4//5)th` of its initial volume. Find the resulting temperature.

Answer

`-25^(@)C`

Explanation

Here, `T_(1) = 37^(@)C = (37+273)K = 310 K`
`V_(2) = 4/5 V_(1) T_(2) = ?`
When pressure remains constant.
`(V_2)/(V_1) = (T_(2))/(T_(1))`
`T_(2)=(V_(2))/(V_(1))xxT_(10)=(4)/(5) xx 310 = 248 K`
`=(248-273)^(@)C = -25^(@)C`