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JEE Main levelSingle correct optionDifficulty 2 of 3sc016

A gas obeys $θ = K p / (1 + K p)$ with $K = 0.50 at m^{- 1}$ . The fractional surface coverage at $p = 8.0 atm$ is:

A $0.20$
B $0.80$
C $0.44$
D $0.50$

Answer: B

K p = 0.50 \times 8.0 = 4.0

, so

θ = 4.0/ (1 + 4.0) = 0.80

. The values

0.20

and

0.44

come from inverting the formula or dividing by

1 + p

.

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