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When an open vessel at 27^(@)C ws heated...

When an open vessel at `27^(@)C` ws heated, three-fifths of the air is in escaped from it. If the volume of the vessel remained unchanged, calculate the temperature at which the vessel was heated?

Text Solution

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Let the amount of air present in the vesse at `27^(@)C` was a mol, and when it was heated to TK, `(3)/(5)`th of the air was expelled.
So, the amount of air in the vessel after heating
`=(n-(3n)/(5))=(2)/(5)`n mol
As the vessel was open and its volume remained unchanged on heating, the pressure (P) and teh volume (V) of the air present in the vessel would be the same as those at `27^(@)C`.
At temperature 300 K: `PV=nRxx300`
At temperature T K : `PV=(2)/(5)nRT`
Thus, `nRxx300=(2)/(5)nRT or , T=750K`
`therefore` The vessel was heated at `(750-273)^(@)C=477^(@)C`.
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