91Ó°ÊÓ

1. The change in internal energy along the path ca is $\mathrm{Δ}{E}_{int}=−160\text{J}$
2. The energy transferred to the gas along the ab is$Q_{ab}=200\text{J}$
3. The energy transferred to the gas along the path bc is $Q_{bc}=40\text{J}$

The first law of thermodynamics states that the change in the internal energy of the system is defined as the difference between the amount of heat produced in the process and the net work done by the system subjected to the conservation of energy. Thus, the heat produced in the process is neither created nor destroyed. We can use the concept of the first law of thermodynamics and also the concept for the constant volume processes, which states there is no change in volume. Hence, there is no work done for a constant volume change process.

Formulae:

From the first law of thermodynamics, $\mathrm{Δ}{E}_{int}=Q−W$ …(¾±)

Where, $\mathrm{Δ}{E}_{int}$is the change in internal energy of the system, Q is the heat produced, W is the work done by the system.

The total work done during the two steps path from a to b and b to c . Along path b to c , there is no change in volume.Hence, there is no work done. Work takes place only along the path a to b .

According to equation (i), the change in the internal energy along the pathis given as:

$\begin{array}{c}\mathrm{Δ}{E}_{int}=E_{inta}−E_{intc}\\ =E_c−E_b+E_b−E_c\\ =Q_{bc}−W_{bc}+Q_{ab}−W_{ab}\text{(fromequation(i))}\\ 160\text{J}=40\text{J}−0\text{J}+200\text{J}−W_{ab}\\ W_{ab}=80\text{J}\end{array}$

The work done by the gas along the path abc is$W_{ab}$

Hence, it is given as:

$\begin{array}{c}{W}_{abc}=W_{ab}\\ =80\text{J}\end{array}$

Hence, the work done by the gas along the path abc is $80\text{J}$

From the calculations of part (a), we can get the work done by the gas.

The work done by the gas along the path ab is $80J$