Why does an isothermal expansion require more energy?

Question

Let's say we have two completely identical cylinders of some ideal gas (same $P,V,T,n$, etc...), and we just want to double each one's volume.

First Cylinder

The first cylinder undergoes a simple reversible isothermal expansion, from $V_i$ to $2V_i$ (and likewise, $P_i \to P_i/2$). The work done is just $$W = P_i V_i \ln \left(V_f/V_i\right) = P_i V_i \ln 2 = nRT_i \ln 2$$ But, since this is an isothermal process, $\Delta U = 0$, and so therefore by the first law, the environment has to supply $Q_1 = W = P_i V_i \ln 2$ to the system in order for this process to occur.

Second cylinder

The second cylinder undergoes two steps:

1. A reversible adiabatic expansion from $V_i \to 2V_i$. So, the work done will be $$W = \frac{P_iV_i^\gamma \left(V_f^{1-\gamma} - V_i^{1-\gamma}\right)}{1-\gamma} = P_i V_i \left(\frac{2^{1-\gamma}-1}{1-\gamma}\right)$$ And, since this is an adiabatic process, the environment transfers no heat/energy, and so the system loses $\Delta U = -W = -P_i V_i \left(\frac{2^{1-\gamma}-1}{1-\gamma}\right)$ from its internal energy to perform this work.

2. A reversible isochoric heating back up to the original initial temperature $T_i$. Since, due to the earlier step, the system just lost $$ \Delta U = -W = -P_i V_i \left(\frac{2^{1-\gamma}-1}{1-\gamma}\right)$$ then all we need to do to get the system back up to its original internal energy/temperature is have the environment supply $Q_2 = P_i V_i \left(\frac{2^{1-\gamma}-1}{1-\gamma}\right)$ back into the system.

Question

In both processes, we have some cylinder migrate from some initial state $\left(P_i,V_i\right)$ to a final state of $\left(P_i/2, 2V_i\right)$. In both processes, the starting and end points are the same, and neither system has a net increase or decrease in its internal energy. So, for both cases, the environment must supply all the energy anyway (in the form of heat) to do everything. But, Mayer's Relation states that $\gamma = C_p/C_v = \left(C_v + R\right)/C_v$, which implies $$Q_1 = P_i V_i \ln 2 \ne Q_2 = P_i V_i \left(\frac{2^{1-\gamma}-1}{1-\gamma}\right)$$

Why is this so? Why does the energy supplied by the environment to expand cylinder 1 have to be different than the energy supplied to expand cylinder 2? There's no net change in either system's internal energy and the starting and end points for each system are identical! So why does the environment have to supply different amounts of energy for one versus the other? Is there a hidden "irreversibility" here that I'm missing? I think all of the processes I mentioned can be done reversibly.

Edit: in fact, in order to make $Q_1 = Q_2$, we must violate Mayer's Relation and have $$\gamma = 1 - \lg \left(1 -\frac{R\ln 2}{C_v}\right) \ne \frac{C_v + R}{C_v}$$ It seems to me that the environment "has to" supply the same amount of energy to both cylinders for the sake of energy conservation (since neither cylinder has a net change in internal energy but goes from identical starting to identical end states). Oddly enough, the entropy transfer for both cylinder 1 (isothermal) and cylinder 2 (adiabatic + isochoric) is identical however if Mayer's Relation is true.

Answer 1

You need more heat in the isothermal expansion because the gas does more work, due to the fact that the pressure is larger than during the adiabatic expansion.