300 L of ammonia gas at `20^(@)C` and 20 atm pressure is allowed to expand in a space of 600 L capacity and to a pressure of 1 atm. Calculate the drop in temperature.

Answer

B, C

Explanation

`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)), i.e., (20 atmxx300 L)/(293 K)=(1 atmxx600 L)/(T_(2))" or " T_(2)=29.3 K`
Drop intemperature =293-29.3=263.7 K