91Ó°ÊÓ

From the given data of the figure,

1. Along path iaf,$\mathrm{Q}=50\mathrm{cal}\mathrm{and}\mathrm{W}=20\mathrm{cal}$
2. Along path ibf,$\mathrm{Q}=36\mathrm{cal}$
3. Along path fi,$\mathrm{W}=-13\mathrm{cal}$
4. ${\mathrm{E}}_{\mathrm{int},\mathrm{i}}=10\mathrm{cal}$
   
5. ${\mathrm{E}}_{\mathrm{int},\mathrm{b}}=22\mathrm{cal}$
   

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. Similarly, the internal energy radiated at constant pressure is zero. Using these properties, we can solve the given.

Formula:

The first law of thermodynamics, $\mathrm{dQ}=\mathrm{dU}+\mathrm{dW}$ …(¾±)

Using equation (i), the value of internal energy can be given as:

$\begin{array}{rcl}\mathrm{dU}& =& 50\mathrm{cal}-20\mathrm{cal}\\ & =& 30\mathrm{cal}\end{array}$

But , along path ibf,$\mathrm{dQ}=36\mathrm{cal}$

And , $\mathrm{dU}=30\mathrm{cal}$, since the internal energy of system only depends on its final and initial position of the body in the process.

The work done along path ibf is given using equation (i) as:

$\begin{array}{rcl}\mathrm{dW}& =& \left(36-30\right)\mathrm{cal}\\ & =& 6\mathrm{cal}\end{array}$

Hence, the required amount of work done is$6\mathrm{cal}$

For the return path fi,

The value of work done is given, that is$\mathrm{dW}=-13\mathrm{cal}$

Now, the internal energy during the process as it depends of the final and initial points and hence for the return path is given as:$\mathrm{dU}=-30\mathrm{cal}$

So, the heat produced in this path fi is given using equation (i) as:

$\begin{array}{rcl}\mathrm{dQ}& =& \left(-30\mathrm{cal}\right)+\left(-13\mathrm{cal}\right)\\ & =& -43\mathrm{cal}\end{array}$

Hence, the value of Q is$-43\mathrm{cal}$

If, the initial internal energy is given as:${\mathrm{E}}_{\mathrm{int},\mathrm{i}}=10\mathrm{cal}$

The internal energy of the system is given as: $\mathrm{dU}=30\mathrm{cal}$

So, the final internal energy is given by:

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{int},\mathrm{f}}& =& 10\mathrm{cal}+30\mathrm{cal}\\ & =& 40\mathrm{cal}\end{array}$

Hence, the value of final internal energy is$40\mathrm{cal}$

Along path bf, at constant volume, the work done is zero.

Then using the given data in equation (i), we can get the heat produced along path bf as follows:

$\begin{array}{rcl}{\mathrm{Q}}_{\mathrm{bf}}& =& {\mathrm{E}}_{\mathrm{int},\mathrm{f}}-{\mathrm{E}}_{\mathrm{int},\mathrm{b}}\\ & =& 40\mathrm{cal}-22\mathrm{cal}\\ & =& 18\mathrm{cal}\end{array}$

Therefore, the heat produced in the process is $18\mathrm{cal}$

Along path ibf, the amount of heat produced is given as:

$\mathrm{Q}=36\mathrm{cal}$

Now, we calculated the heat produced along path bf as:${\mathrm{Q}}_{\mathrm{bf}}=18\mathrm{cal}$

So, the amount of heat produced along path ib can be given as:

$\begin{array}{rcl}{\mathrm{Q}}_{\mathrm{ib}}& =& \mathrm{Q}-{\mathrm{Q}}_{\mathrm{bf}}\\ & =& 36\mathrm{cal}-18\mathrm{cal}\\ & =& 18\mathrm{cal}\end{array}$

Therefore, the heat produced in the process is$18\mathrm{cal}$.