A gas within a closed chamber undergoes the cycle shown in the p-V diagram of Figure. The horizontal scale is set by${\mathit{V}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.0}{\mathbf{\text{m}}}^{\mathbf{\text{3}}}$. Calculate the net energy added to the system as heat during one complete cycle.

From thegiven figure, we have the volume as

i)$V_A=V_c=1{\text{\hspace{0.17em}m}}^{\text{3}}$

ii)$V_B=4{\text{\hspace{0.17em}m}}^{\text{3}}$

And pressure as

iii)$p_B=p_C=30\text{\hspace{0.17em}Pa}$

iv)$p_A=10\text{\hspace{0.17em}Pa}$

We have the integral $\mathit{W}\mathbf{=}\underset{}{\overset{}{\int }}\mathit{p}\mathbf{.}\mathit{d}\mathit{V}$

i.For an isochoric process, pressure$p$is constant; then,$\mathit{W}\mathbf{=}\mathit{p}\mathit{\Delta }\mathit{V}$

ii. For an isochoric process,is constant; therefore, we get$\mathit{W}\mathbf{=}\mathbf{0}\mathbf{.}$

The internal energy is the same over the entire cycle, so the heat energy absorbed must be equal to the work done:

$W=Q.$

Over the path $\mathit{A}$to $\mathit{B}$the pressure is a linear function of volume, so we write.

$\mathit{p}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathit{V}$

Then the work done

role="math" localid="1661839902618" ${\mathit{W}}_{\mathbf{AB}}\mathbf{=}\underset{{\mathbf{V}}_{\mathbf{A}}}{\overset{{\mathbf{V}}_{\mathbf{B}}}{\int }}\mathit{p}\mathit{d}\mathit{V}$

During BC cycle, pressure is constant. Then,${\mathit{W}}_{\mathbf{B}\mathbf{C}}\mathbf{=}{\mathit{p}}_{\mathbf{B}}\mathit{\Delta }{\mathit{V}}_{\mathbf{B}\mathbf{C}}$

During the CA cycle, volume is constant, so no work is done during the process, so${\mathit{W}}_{\mathbf{C}\mathbf{A}}\mathbf{=}\mathbf{0}$

Formulae:

The work done by the heat system during the path from A to reach B,

$W_{AB}=\underset{V_A}{\overset{V_B}{}}pdV$ …(i)

The total work done, $W=W_{AB}+W_{BC}+W_{CA}$ …(ii)

The work done when pressure is constant, $W=p\Delta V$ …(iii)

Over the path. $A$to$B$the pressure is a linear function of volume, so we write the pressure expression as:

$p=a+bV$ …(iv)

Then, the work done using equation (i) is given as:

.$W_{AB}=\underset{V_A}{\overset{V_B}{}}(a+bV)dV$

$W_{AB}=a(V_B-V_A)+\frac{1}{2}b(V_B^2-V_A^2)$ …(v)

First, we have to find the values of$a\text{\hspace{0.17em}and\hspace{0.17em}}b$. From the figure$p=10\text{\hspace{0.17em}Pa}\text{when}v=1.0{\text{\hspace{0.17em}m}}^3$,and.$p=30\text{\hspace{0.17em}Pawhen}v=4.0{\text{\hspace{0.17em}m}}^{\text{3}}$ Thus, we have equation (iv) as

Then, solving these two equations, we have

$a=\frac{10}{3}\text{\hspace{0.17em}Pa}$and$b=\frac{20}{3}{\text{\hspace{0.17em}Pa/m}}^{\text{3}}$.

Substituting these values and the given values, we get the form of equation (a) as:

$\begin{array}{c}{W}_{AB}=\left(\frac{10}{3}\text{Pa}\right)(4.0{\text{\hspace{0.17em}m}}^{\text{3}}-1.0{\text{\hspace{0.17em}m}}^{\text{3}})+\frac{1}{2}\left(\frac{20}{3}{\text{\hspace{0.17em}Pa/m}}^{\text{3}}\right)[{\left(4.0{\text{\hspace{0.17em}m}}^{\text{3}}\right)}^2-{\left(1.0{\text{\hspace{0.17em}m}}^{\text{3}}\right)}^2]\\ =10\text{\hspace{0.17em}J}+\frac{150}{3}\text{\hspace{0.17em}J}\\ =60\text{\hspace{0.17em}J}\end{array}$

During BC cycle, pressure is constant; then work done using equation (iii) is given as:

$\begin{array}{c}{W}_{BC}=p_B(V_C-V_B)\\ =30\text{\hspace{0.17em}Pa}(1.0{\text{\hspace{0.17em}m}}^{\text{3}}-4.0{\text{\hspace{0.17em}m}}^{\text{3}})\\ =-90\text{\hspace{0.17em}J}\end{array}$

During the CA cycle, volume is constant, so no work is done during the process, so

$W_{CA}=0$

The total work done using equation (ii) is given as:

$\begin{array}{c}W=60\text{\hspace{0.17em}J}-90\text{\hspace{0.17em}J}+0\\ =-30\text{\hspace{0.17em}J}\end{array}$

Thus, the total heat absorbed is$Q=W=-30\text{\hspace{0.17em}J}$

Thus, this means that the gas loses of$30\text{\hspace{0.17em}J}$ energy in the form of heat.