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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» `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))=(20" atm"xx300L)/(293K)=(1" atm"xx600L)/(T_(2))` or `T_(2)=29.3K` `therefore` drop in temperature `=293-29.3=263.7K` |
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