1.

For isothermal expansion of an ideal gas, the correct combination of the thermodynamic parameters will be

Answer»

`DELTA U =0, Q = 0 , w ne 0` and ` Delta H ne 0`
`Delta U ne 0 , Q ne 0 , w ne 0` and `Delta H = 0`
`Delta U = 0 , Q ne 0 , w = 0 ` and ` Delta H ne 0`
`Delta U = 0 , Q ne 0 , w ne 0` and `Delta H = 0`

Solution :For ISOTHERMAL PROCESS , ` Delta T= 0`
` THEREFORE Delta U = nC_v Delta T= 0`
` DeltaH = nC_p Delta T = 0`
From first LAW of thermodynamics
` Delta U = Q + w`
As ` Delta U = 0`
` therefore Q = w ne 0`


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