A gaseous reaction, 3A --> 2B; is carried out in a 0.0821 litre closed container initially containing 1 mole of gas A. After sufficient time a curve of P (atm) vs T (K) is plotted and the angle with x-axis was found to be 42.95°. The degree of association of gas A is ?[Given : tan 42.95 = 0.8]
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Explanation:
Slope = tan θ= tan 42.95=0.8=VnR
n=0.08210.8×0.0821=0.8
A
B
Initial number of moles
1
0
Equilibrium number of moles
1−3x
2x
Total number of moles at equilibrium =1−3x+2x=1−x=0.8
Hence, x=0.2
The degree of association is 13×0.2=0.6