T C said:
In the first scenario, I have calculated the net energy consumptioOn by subtracting the work done by the gas from the energy spent by the piston i.e. (4539.38 - 2240.28) J. And now, as per you, it should be (4539.38 + 2240.28) J, right?
There is no net work done on the piston, and, thus, no net energy "spent by the piston." I show this below.
From Newton's 2nd law of. motion applied to the piston, we have $$P_{gas}A+F-P_{atm}A=0\tag{1}$$The term ##P_{gas}A## represents the forward force of the gas on the inside face of the piston, the term F represents the forward external pulling force (needed to advance the piston) applied on the outside face of the piston, and the term ##P_{atm}A## represents the backward pressure force exerted by the air on the outside face of the piston. Notice that, since we are assuming a massless frictionless piston, the net force on the piston is zero at all times during the process. This means that the net of the forces exerted on the faces of the piston is zero.
Next let's consider the work done on the piston (and the net energy consumed by the piston). If we multiply Eqn. 1 by the differential forward displacement of the piston ds and integrate between the Initial and final states of the system, we obtain: $$\int_{V_i}^{V_f}{P_{gas}dV}+\int{Fds}-P_{atm}(V_f-V_i)=0\tag{2}$$The first term on the left-hand side represents the work done by the gas on the inside face of the piston; the second term represents the work done by the applied external force on the outside face of the piston; and the third term (with minus sign included) represents minus the work done by the atmosphere on the outside face of the piston. Eqn. 2 shows that (1) the net work done
on the piston
by its surroundings is zero and the net work
by the piston
on its surroundings is zero. Therefore, with regard to the piston, advancing it consumes no net energy.
Let us rearrange Eqn. 2 to obtain:$$\int_{V_i}^{V_f}{P_{gas}dV}=P_{atm}(V_f-V_i)-\int{Fds}\tag{3}$$The term on the LHS represents the work done by the gas on its surroundings (the inside piston face). The LHS represents the work done by the outside face of the piston on the surroundings of the combined system comprising the gas and the piston. You can see that the work done by the gas on its surrounding is exactly equal to the work done by the combined system of gas and piston on its surroundings.
When applying the first law of thermodynamics to a process, the first step in the analysis must be to define, in a precise manner, the specific system that is being analyzed. In the case of scenario 1, there are two simple choices one can make:
1. The gas alone
2. The combination of the gas and the cylinder
Analyzing the process based on either of these choices gives exactly the same results,
CHOICE 1: System = gas alone
In this case, the first law of thermodynamics gives, $$\Delta U=nC_v(T_f-T_i)=-\int_{V_i}^{V_f}{P_{gas}dV}\tag{4}$$
CHOICE 2: System = gas plus piston
In analyzing this case, it is necessary to assume that the massless frictionless piston is thermally insulated such that the internal energy change of the piston is zero. (This is an implicit assumption in case 1 as well). If we write down the equation for the first law of thermodynamics in this case, we obtain $$\Delta U=nC_v(T_f-T_i)=-P_{atm}(V_f-V_i)+\int{Fds}\tag{5}$$Next, combining this with Eqn. 3, we obtain:$$\Delta U=nC_v(T_f-T_i)=-\int_{V_i}^{V_f}{P_{gas}dV}\tag{6}$$
Note that Eqns. 5 and 6 are, as expected, identical..
We have shown here that, for scenario 1,
- No net energy is consumed in advancing the piston from its initial position to its final position (i.e., no net work is done on the piston)
- The only energy consumption is the result of expanding the gas. However, even here, no energy is dissipated because the decrease in internal energy of the gas is exactly equal to the work done on the surroundings (the work to push the atmosphere back minus the work expended in using an external tensile force F to pull the piston).