1.

One mole of an ideal gas is allowed to expand reversibly and adibatically from a temperature of 27^(@)C. If the work done during the process is 3 kJ, then final temperature of the gas is (C_(v)=20 J//K)

Answer»

100 K
150 K
195 K
255 K

Solution :Given nubmer of moles =1
Initial temperature = `27^(@)C=300K`
Work DONE by the SYSTEM = 3kJ = 3000 J
It will be (-) because work is done by the system
HEAT capacity at constant volume (CV) = 20 J/K
We know that work done
`W=-nC_(V)(T_(2)-T_(1)), 3000 =-1xx20(T_(2)-300)`
`3000=-20T_(2)+6000`
`20T_(2)=3000, T_(2)=(3000)/(20)=150 K`


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