Answer:
Short answer: because of the expansion of the gas inside requires energy to push against the atmospheric pressure in the constant pressure case. And this doesn't happen in the constant volume case
Explanation:
Let's assume ideal gas behaviour, we know that the internal energy of an ideal gas is function of its' temperature only and viceversa. Let's look at the constant volume case.
If heat is added to a gas at constant volume, all the heat is stored by the system, i.e. all the heat becomes internal energy, because no work is done ( fixed volume, no expansion work possible)
[tex]d U = \delta Q +\delta W \\\delta W=0\\d U = \delta Q[/tex]
Therefore, all the heat absorbed by the system ends up raising the temperature of the gas.
Now, in the case of the constant pressure system, we see that:
[tex]PV=nRT\\T=\frac{PV}{nR}[/tex]
[tex]T= \frac{P}{nR}V[/tex]
where [tex]\frac{P}{nR}[/tex] remains constant. Thus if T increases, V must also increase, that means, the gas must expand.
So, if the gas receives the same amount of heat as the constant volume system, the first law balance will look like:
[tex]d U = \delta Q+ \delta W\\d U = \delta Q - P\cdot dV[/tex]
where [tex]P\cdot dV[/tex] is the expansion work. We see then that [tex]dU[/tex] will be higher for the constant volume case than for the constant pressure case, if we give the system the same [tex]\delta Q[/tex] amount of heat, because some of the energy that gets in in form of heat will be lost by expansion work done to keep the pressure constant.
