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• Pressure${\text{P}}_{ab}=5\text{鈥塳笔补}$
• Pressure${\text{P}}_c=2\text{鈥塳笔补}$
• Volume${\text{V}}_{bc}=4{\text{鈥尘}}^3$
• Volume${\text{V}}_a=2{\text{鈥尘}}^3$

The internal energy of an ideal gas is given by,

$螖E_{int}=\left(\frac{f}{2}\right)pV$

Here$f$ is the degree of freedom,$f$ is the pressure,$V$ is the volume.

Substitute $nRT$for$pV$ into the above equation.

$螖E_{int}=\left(\frac{f}{2}\right)nRT$

Here$n$ is the number of moles, $R$is the gas constant,$T$ is the temperature.

The equipartition of energy theorem states that each degree of freedom of a molecule has associated with it an energy of $\frac{1}{2}RT$per mole. If$f$ is the number of degree of freedom, then${\text{E}}_{int}=\left(\frac{f}{2}\right)nRT$

Usingtheideal gas equation, we get

$\begin{array}{c}pV=nRT\\ E_{int}=\left(\frac{f}{2}\right)pV\end{array}$

For a diatomic gas, the degree of freedom$f=5$

Change in the internal energy of an ideal diatomic gas:

The internal energy at point$a$is

$E_{int}=\left(\frac{5}{2}\right)p_a{V}_a$

And internal energy for point$c$

$E_{int}=\left(\frac{5}{2}\right)p_c{V}_c$

The change in internal energy is then

$\begin{array}{c}{\text{螖贰}}_{int}=\left(\frac{5}{2}\right)p_c{V}_c鈭�\left(\frac{5}{2}\right)p_a{V}_a\\ =\left(\frac{5}{2}\right)(p_c{V}_c-p_a{V}_a)\\ =\left(\frac{5}{2}\right)((2脳{10}^3\text{Pa})\left(4{\text{鈥尘}}^3\right)鈭�(5脳{10}^3\text{Pa})\left(2{\text{鈥尘}}^3\right))\\ =鈭�5脳{10}^3\text{J}\end{array}$

Therefore the change in the internal energy of an ideal diatomic gas for path$ac$ is $鈭�5.0脳{10}^3\text{J}$.

Energy added to the gas as heat for path$ac$:

The work done along the path$ac$ is

$\begin{array}{c}{W}_{ac}=\left(\frac{P_a+P_c}{2}\right)(V_c-V_a)\\ =\left(\frac{5+2}{2}\right)(4鈭�2)\\ =(3.05脳{10}^3\text{Pa})\left(2{\text{鈥尘}}^3\right)\\ =7脳{10}^3\text{J}\end{array}$

The first law of thermodynamics gives

$\begin{array}{c}{Q}_{ac}=螖E_{int}+W_{ac}\\ =(鈭�5脳{10}^3+7脳{10}^3)\text{J}\\ =2脳{10}^3\text{J}\end{array}$

Therefore the energy added as heat for path$ac$ is $2.0脳{10}^3\text{J}$.

Heat required if the gas goes from$a$to$c$ :

$螖{\text{E}}_{int}$depends only on the initial and final states and not on the 鈥減ath鈥� between them.

Then we can write this for indirect path:

$\begin{array}{c}螖E_{int}=E_{int,c}鈭�E_{int,a}\\ =鈭�5脳{10}^3\text{J}\end{array}$

In this case, the total work done consists of the work done during the constant pressure part (the horizontal line in the graph) plus the work done during the constant volume part (the vertical line).

$\begin{array}{c}{W}_{indirect}=(5脳{10}^3\text{Pa})\left(2{\text{鈥尘}}^3\right)+0\\ =(5脳{10}^3\text{Pa})\left(2{\text{鈥尘}}^3\right)+0\\ =1脳{10}^4\text{J}\end{array}$

Now, from first law of thermodynamics,

$\begin{array}{c}{\text{Q}}_{indirect}={\text{螖贰}}_{int}+{\text{W}}_{indirect}\\ ={\text{螖贰}}_{int}+{\text{W}}_{indirect}\\ =(鈭�5脳{10}^3)+(1脳{10}^4)\\ =5脳{10}^3\text{J}\end{array}$

Therefore theHeat required if the gas goes from$a$to$c$along the path$abc$is$5.0脳{10}^3\text{J}$ .