What is the heat needed for an isothermal expansion of gas. Why do physics and chemistry yield different answers? One mole of a certain ideal gas is contained under a weightless piston of a vertical cylinder at a temperature T. The face of the piston opens into the atmosphere. What is the heat supplied in the process to expand the gas from volume $V_1$ to $V_2$ isothermally ? Friction of the piston against the cylindrical wall is negligible is small


Now thermodynamics is a common topic to both physics and chemistry and as per my understanding of thermodynamics I am getting different results for the heat that will need to be supplied.

Chemistry says $$∆U =∆Q +∆W $$ and defines work as the work done on the gas and this will give $$W = -P_{atm}(V_2-V_1) $$ and as for isothermal process$$ ∆U =0 , ∆Q =P_{atm}(V_2-V_1)$$

According to physics on the other hand $$∆Q =∆U+∆W$$ and here the work done is the work done by the gas. That is 

$$W = \int_{V_1}^{V_2} PdV=\int_{V_1}^{V_2} \frac{RT}{V}dV = RT \ln V_2-RT \ln V_1$$ and as for isothermal process $$∆U =0 ,\, ∆Q =RT \ln \left(\frac{V_2}{V_1}\right)$$ 

I am also uncertain that maybe neither of these is the correct expression and the network ok is actually in that work from both these forces and we have to take the sum of these to yield $$∆Q =RT\ln \left(\frac{V_2}{V_1}\right)-P_{atm}(V_2-V_1)$$

Now the value of Q should be unique because in the real world when we perform an experiment, only one value of Q would be supplied.

So which one of these is correct and where am I lacking in my understanding of thermodynamics from a chemistry  standpoint ??

Now as this question links both physics and chemistry, I want to post it on both sites (I hope it's fine) and the link to it is here
The problem is that both the problem statement you presented and the chemistry solution you showed are fallacious.  The fallacy is that, for an ideal gas $V=nRT/P$, so that, if the pressure and temperature are both constant, the volume can't change.  If the volume does change, then either the pressure must change or the temperature must change.  

In the "chemistry solution" you presented, the pressure is taken as constant (atmospheric).  So the temperature must change.  But if the temperature changes, then $\Delta U$ is not zero (as they assume).  So the solution is incorrect.  The correct solution, based on the 1st law, is $$\Delta U=nC_v\Delta T=Q-P_{atm}(V_2-V_1)$$or equivalently, $$\Delta H=\Delta U+P_{atm}(V_2-V_1)=nC_p(T_2-T_1)=Q$$Also, since $$n=\frac{P_{atm}V_1}{RT_1}$$we have $$\frac{T_2}{T_1}=1+\frac{QR}{C_pP_{atm}V_1}$$

In the "physics solution" you presented, they make the opposite assumption that, rather than the pressure remaining constant, the temperature remains constant.  For the temperature remaining constant, the solution you presented is correct.  But the final pressure is:  $$P_2=P_{atm}\frac{V_1}{V_2}$$