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Gas A (1 mol) dissociates in a closed rigid container of volume 0.16 lit as per following reaction.

$$\ce{2A (g) -> 3B (g) + 2C (g)}$$

If degree of dissociation of $\ce{A}$ is 0.4 and remains constant in entire range of temperature, then correct P Vs T graph is

  1. Slope 0.75 and passing through origin
  2. Slope 0.8 and passing through origin
  3. slope 1.1 and passing through origin
  4. slope 0.4 and passing through origin

\begin{align} \ce{2A &&-> 3B& + &2C}\\ 1-2x&&3x&&2x\\ \end{align}

where $2x$ is dissociation constant. Substituting value of $x$ we get moles of A, B and C to be 0.4, 0.6 and 0.4 respectively.

I am confused about how to proceed after this. How to proceed?

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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.

$$\ce{2A -> 3B + 2C}$$

or

$$\ce{A -> }\dfrac{3}{2}\ce{B + C}$$

let B and C equal mythical gas X and

$$\ce{A -> }\dfrac{5}{2}\ce{X}$$

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.

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  • $\begingroup$ The units of R is indeed given in 0.08litol/K.Sorry for not including this and thanks for your answer $\endgroup$ – Scáthach Jun 23 '18 at 9:03
  • $\begingroup$ So even though the gases are reacting,we can consider them to be one ideal gas at equilibrium point .Can we apply ideal gases individually at the equilibrium for the gases and calculate pressure but that would be go against Dalton's partial pressure law ,right? $\endgroup$ – Scáthach Jun 23 '18 at 9:54
  • $\begingroup$ Also at eqiluilibrium even though the reaction still goes on ,we can apply the gas law......is there a reason? $\endgroup$ – Scáthach Jun 23 '18 at 9:55
  • $\begingroup$ Can you comment on doubts pls as it will be very much helpful for me $\endgroup$ – Scáthach Jun 25 '18 at 1:52
  • $\begingroup$ Yes, you could calculate the individual pressures. It is just a bit more math. // You actually drew the reaction arrow wrong. You have it only going to the right, when the arrows should go both ways since there is an equilibrium (in other words both forward and reverse reactions). $$\ce{2A (g) <=> 3B (g) + 2C (g)}$$ $\endgroup$ – MaxW Jun 25 '18 at 2:31

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