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Q. On isothermal expansion of ideal gas, its

Rajasthan PETRajasthan PET 2007

Solution:

Isothermal process $ dT=0, $
$ \Delta E=0 $ $ H=E+pV $
$ \Delta H=\Delta E+p.\Delta V $
where, $ \Delta H= $ change in enthalpy
$ \Delta E= $ change in internal energy
$ p= $ pressure
$ \Delta V= $ change in volume during expansion of ideal gas
If $ pV=nRT $
$ \Delta H=\Delta E+\Delta (nRT) $
$ \Delta H=\Delta E+nR.\Delta T $
$ \Delta T=0 $ $ \Delta E=0 $ (Isothermal expansion)
Hence, $ \Delta H=0 $
During isothermal expansion, the enthalpy of ideal gas remains unchanged.