5 g of unknown gas has pressure P at a temperature T K in a vessel. On increasing the temperature by `50^(@)C`, 1 g of the gas was given out to maintain the pressure P. The original temperature was :

Explanation

Initially, `T_(1)=T,P_(1)=P,V_(1)=V `(Volume of the vessel), `n_(1)=(5)/(M)`mole
Finally, `T_(2)=(T+50)K, P_(2)=P, V_(2)=V, n_(2)=(5-1)/(M)=(4)/(M)` mole
(Increase of `50^(@)C`=increase of 50 K because size of `1^(@)C`=1 K)
Applying PV=nRT
`n_(1)T_(1)=n_(2)T_(2)`
`(5)/(M)xxT=(4)/(M)(T+50)" or "5T=4T+200" or T=200 K`