Finding equation of line of the P-T graph for a gas dissociation reaction
The overall notion is that gases A, B and C are all behaving as an ideal gases which can be represented by the mythical ideal gas X. Then the moles of X are equal to moles of A+B+C. Then we can use the ideal gas equation:

$$pV=nRT$$

which simplifies to:

$$p = \dfrac{nR}{V}\times T$$

Given that the degree of dissociation of A is 0.4 and we started with 1 moles of A, then 0.6 moles of A remain and 0.4 moles got converted. 

$$\ce{2A -&gt; 3B + 2C}$$

or 

$$\ce{A -&gt; }\dfrac{3}{2}\ce{B + C}$$

let B and C equal mythical gas X and 

$$\ce{A -&gt; }\dfrac{5}{2}\ce{X}$$

but since A = 0.6 moles of X, the total moles are 

$$n = 0.6 + 0.4\times\dfrac{5}{2} = 1.6\text{ moles}$$

We know that n = 1.6 moles and that V = 0.16 L so:

$$p =\left(\dfrac{n}{V}\right)RT = \dfrac{1.6\text{ mole}}{0.16\text{liter}}RT = 10RT\dfrac{\text{moles}}{\text{liter}}$$

Now the problem doesn't really specify what value of the ideal gas constant, $R$, to use nor what units the temperature, $T$, should be. 

The temperature must be an absolute scale since the line passes through the origin. Thus of the main choices of Kelvin and Rankine, using Kelvin seems reasonable for temperature. 

but there are still options for the pressure units of R


(a) $8.314\text{ L}\cdot\text{kPa}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(b) $8.314\times10^{−2} \text{ L}\cdot\text{bar}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(c) $62.36 \text{ L}\cdot\text{torr}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(d) $8.206\times10^{−2}\text{ L}\cdot\text{atm}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$


For $10R$ to be between 0.4 and 1.1 the only answers that fit are (b) and (d). Thus the pressure could be either in atm or bar. Presumably the OP's book would have stated a preference for R.