Two moles of an ideal gas is expanded isothermally and reversibly from `1` liter to `10` liter at `300 K`. The enthalpy change (in `kJ`) for the process

Options

A. 11.4 kJ
B. `-11.4kJ`
C. `0 kJ`
D. `4.8 kJ`

Answer

c

Explanation

`DeltaH =Delta( U + PV) = DeltaU + Delta(RT) =DeltaU + R DeltaT`
For isothermal expansion of ideal gas, `DeltaU = 0`
`( :' DeltaU =C_(v) DeltaT ` and `DeltaT =0)`
`:. DeltaH =0 +0, i.e., DeltaH =0`
or putting `DeltaU = C_(v) DeltaT , DeltaH = C_(v) DeltaT + R DeltaT`
`= (C_9v) +R) DeltaT = C_(p) DeltaT` for 1 mole.
For n moles, `DeltaH = n C_(p) DletaT =0`