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

Options

A. `Delta U = 0, Q = 0, w ne 0 and Delta H ne 0`
B. `Delta U ne 0, Q ne 0, w ne 0 and Delta H ne 0`
C. `Delta U= 0, Q ne 0, w= 0 and Delta H ne 0`
D. `Delta U = 0, Q ne 0, w ne 0 and Delta H =0`

Answer

D

Explanation

For isothermal process, `Delta T = 0`. From first law of thermodynamics `Delta U= Q + W`
`therefore Delta U= nC_(v) Delta T= 0 " As " Delta U= 0`
`Delta H= nC_(P) Delta T= 0 " " therefore Q= W ne 0`