Gas A ( 1 mol ) dissociates in a closed rigid container of volume 0.16 lit. as per following reaction.
$2 A (g) ⟶ 3 B (g) + 2 C (g)$

If degree of dissociation of A is 0.4 and remains constant in entire range of temperature, then the correct P vs T graph is [Given
$R = 0.08$
lit-
$atm / mol / K$
]

Question

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

2 A (g) ⟶ 3 B (g) + 2 C (g)

If degree of dissociation of A is 0.4 and remains constant in entire range of temperature, then the correct P vs T graph is [Given

R = 0.08

lit-

atm / mol / K

]

TestHub · Accepted Answer

Here's a brief solution to the problem:

The reaction is:

$2 A (g) ⟶ 3 B (g) + 2 C (g)$

Initial moles of A = 1

Degree of dissociation, $α = 0.4$

Moles of A reacted = $1 \times 0.4 = 0.4$

At equilibrium:

Moles of A = $1 - 0.4 = 0.6$

Moles of B = $3/2 \times 0.4 = 0.6$

Moles of C = $2/2 \times 0.4 = 0.4$

Total moles at equilibrium, $n = 0.6 + 0.6 + 0.4 = 1.6$

Using the ideal gas equation, $P V = n RT$

$P = \frac{n RT}{V}$

Since the volume $V$ and $R$ are constants, and the degree of dissociation $α$ is constant (so $n$ is constant), we have:

$P \propto T$

Therefore, the graph of P vs T is a straight line passing through the origin.

$P = \frac{1.6 \times 0.08 \times T}{0.16} = 0.8 T$

The correct graph should be a straight line with a slope of 0.8. Option B is the closest to this description.

Question with Solution