Figure represents a closed cycle for a gas (the figure is not drawn to scale). The change in the internal energy of the gas as it moves from a to c along the path abc is$\mathbf{-}\mathbf{200}\mathbf{\text{\hspace{0.17em}J}}$. As it moves from c to d, $\mathbf{180}\mathbf{}\mathbf{\text{J}}$must be transferred to it as heat. An additional transfer of$\mathbf{80}\mathbf{}\mathbf{\text{J}}$to it as heat is needed as it moves from d to a. How much work is done on the gas as it moves from c to d?

a) The internal energy of the gas from a to c during path abc,$\text{Δ}{E}_{int,abc}\text{(Δ}{E}_{int,ac})=-200\text{J}$

b) The energy transferred to heat from c to d,$Q_{cd}=180\text{J}$

c) The energy transferred to heat from d to a,$Q_{da}=80\text{J}$

In this whole thermodynamic cycle, the use of first law of thermodynamics gives the heat produced, internal energy and the work done along a particular path by the body. The work done during a process of constant volume is zero. By using the first law of thermodynamics, we can calculate the work done along the path cd.

Formula:

The equation of the first law of thermodynamics, $Q=\text{Δ}{E}_{int}+W$ …(i)

There is no work done in the process going from$d$to$a$. Hence, the heat transferred to heat during the process using equation (i) is given as:

$\begin{array}{c}{Q}_{da}=\text{Δ}{E}_{int,da}\\ =80\text{J}\end{array}$

Also, since the total change in internal energy around the cycle is zero,

$\begin{array}{c}\text{Δ}{E}_{int,ac}+\text{Δ}{E}_{int,cd}+\text{Δ}{E}_{int,da}=0\\ -200\text{J}+\text{Δ}{E}_{int,cd}+80\text{J}=0\\ \text{Δ}{E}_{int,cd}=120\text{J}\end{array}$

Thus, applying the first law of thermodynamics, the work done on the gas as it moves from$c$to$d$using equation (i) is given as:

$\begin{array}{c}{W}_{cd}=Q_{cd}-\text{Δ}{E}_{int,cd}\\ =180\text{J}-120\text{J}\\ =60\text{J}\end{array}$

Hence, the work done during process cd is$60\text{J}$