Question

In an adsorption experiment, a graph between log( x/m ) versus log p is found to be linear with slope of 45° . The intercept on log(x/m) axis was found to 0.3010. The amount of the gas adsorbed per gram of charcoal under the pressure of 0.5 atm will be:

Answer 1

AnswEr :

From the Question,

• Intercept of (x/m) = 0.301
• Pressure = 0.5 atm
• Angle made by the slope = 45°

According to Freundlich's Isotherm,

$\sf \: \dbinom{x}{m} = k {p}^{ \frac{1}{n}}---------(1)$

Taking log on both sides,

$\longrightarrow \sf \: log \dbinom{x}{m} = log(k) + {log(p}^{ \frac{1}{n} } ) \\ \\ \longrightarrow \sf \: log \dbinom{x}{m} = log(k) + \dfrac{1}{n} {log(p})$

Now, comparing with y = mx + c

$\sf \: c = log(k) \\ \sf \: x = log(p) \\ \sf \: m = \dfrac{1}{n}$

Now,

$\sf \: c = log(k) \\ \\ \implies \sf \: antilog(c) = k \\ \\ \implies \sf k = antilog(0.3010) \\ \\ \implies \boxed{ \boxed{ \sf \: k = 2}}$

Also,

$\sf \: m = \dfrac{1}{n} \\ \\ \implies \sf \: tan45 \degree = \dfrac{1}{n} \\ \\ \implies \boxed{ \boxed{\sf \: n = 1}}$

Amount of adsorption :

$\sf \dbinom{x}{m} = 2 \times (0.5) {}^{1} \ \ \ \ \ \ | Using (1) | \\ \\ \longrightarrow \sf \: \dbinom{x}{m} = 2 \times 0.5 \\ \\ \longrightarrow \boxed{ \boxed{ \sf \dbinom{x}{m} = 1}}$

Answer 2

Answer:

i hope it will help u alot......