Calculate Q,W,DeltaE and DeltaH for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of1.0 bar to a final pressure of0.1 bar at a constant temperature of273 K.

Calculate Q,W,DeltaE and DeltaH for the isothermal reversible expansio

1.	Calculate Q,W,DeltaE and DeltaH for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of1.0 bar to a final pressure of0.1 bar at a constant temperature of273 K.
Answer» Solution :For isothermal reversible expansion of an ideal gas, `W= - 2.303 nRT LOG. ( P_(1))/( P_(2)) = -2.303 XX1 xx8.314 xx273 log. (1)/( 0.1) = - 5227 J` `DeltaE=Q+W`. But`DeltaE =0` for isothermal expansion of ideal gas. Hence, `Q= -W = 5527J` `H= E +PV` or `DeltaH = DeltaE +P DeltaV = DeltaE +P DeltanRDelta T = 0+0=0``( :' T =`constant so that`DELTAT =0)`