1.

Three moles of an ideal gas at `127^(@)C` expands isothermally untill the volume is doubled. Calculate the amount of work done and heat absorbed.

Answer» Correct Answer - `6912 J, 6912 J`
Here, `n=3, V_(2)= 2V_(1), dW= ?, dQ=?`
As the process is isothermal, therefore,
`dW= nRT "log"_(e)((V_(2))/(V_(1)))`
`=3xx8.31(127+273)log_(e)2`
`=3xx8.31xx400xx0.6931= 6912J`
`dQ= dU+dW`
`= 0+6912= 6912J`


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