Remember, your expansion is isobaric. Now you should have had a -ve work answer because the system is losing energy by doing expansion work it's just that the heat it is receiving is positive and greater in magnitude than the work it does. One thing to note is that your work calculation is wrong because on integrating you're removing T from the integral which is incorrect. T increases in this case so you can't get -nRTlnV2/V1. If you really want to know the work term, since P is constant, simply plug the initial or final (T,V) values into the equation of state P=nRT/V which will effectively give you the external pressure (you said this process was reversible) and multiply by the delta V (but put a negative sign in front of that). Now you need the heat term, Q. which under isobaric conditions is equal to delta H. which equals Cp(T2-T1). This value will be positive and should be greater in magnitude than the -ve work term you get. Add the 2 together and voila, you'll get your positive delta E answer which make sense since the internal energy of an ideal gas is solely a function of temp. and the temp. of your gas must go up since P = nRT/V, V is going up and P is not changing.