The graph between logxmv-s log P is straight line inclined at an angle 45- with intercept 0-30- What will be rate of adsorption at pressure 0-4 atm-

The correct option is C 0.8 1n=tan45^∘=1 Log K = 0.3010⇒ K=2,p=0.4 atm xm=K_p^1n=2×(0.4)^1=0.8 #160; ...

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Standard XII

Chemistry

Fruendlich Isotherm

The graph bet...

Question

The graph between $log \frac{x}{m} v / s log P$ is straight line inclined at an angle $45^{\circ}$ with intercept $0.30$ . What will be rate of adsorption at pressure $0.4 a t m$ ?

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Solution

The correct option is C $0.8$
$\frac{1}{n} = tan 45^{\circ} = 1$
Log K = $0.3010 \Rightarrow K = 2, p = 0.4 a t m$
$\frac{x}{m} = K_{p}^{\frac{1}{n}} = 2 \times (0.4)^{1} = 0.8$

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The graph between logxmv/slogP is straight line inclined at an angle 45∘ with intercept 0.30. What will be rate of adsorption at pressure 0.4 atm?

The correct option is C 0.81n=tan45∘=1Log K = 0.3010⇒K=2,p=0.4atmxm=K1np=2×(0.4)1=0.8

The graph between $log \frac{x}{m} v / s log P$ is straight line inclined at an angle $45^{\circ}$ with intercept $0.30$ . What will be rate of adsorption at pressure $0.4 a t m$ ?

The correct option is C $0.8$
$\frac{1}{n} = tan 45^{\circ} = 1$
Log K = $0.3010 \Rightarrow K = 2, p = 0.4 a t m$
$\frac{x}{m} = K_{p}^{\frac{1}{n}} = 2 \times (0.4)^{1} = 0.8$