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CHXII05:SURFACE CHEMISTRY

320818 The slope of the straight line graph between $log \frac{x}{m} and logP$ for the adsorption of a gas on solid is

1

k

2

\log k

3

n

4

\frac{1}{n}

Explanation:

$log(x/m) = (1/n)logP + logk y = mX + C$

CHXII05:SURFACE CHEMISTRY

320819 According to Freundlich adsorption isotherm, at high pressure, the value of $\frac{x}{m}$ is

1 Directly proportional to pressure

2 Inversely proportional to pressure

3 Directly proportional to square of pressure

4 Independent of pressure

Explanation:

$\frac{x}{m}$ doesn't depend on pressure.

CHXII05:SURFACE CHEMISTRY

320820 Which of the following graphs would yield a straight line?

1

\frac{x}{m} V s P

2

l o g (\frac{x}{m}) V s P

3

\frac{x}{m} V s l o g P

4

l o g (\frac{x}{m}) V s l o g P

Explanation:

According to Freundlich adsorption isotherm, $\log (x / m) = (1 / n) \log P + \log k$
$y = m x + C$
Plot of $\log (x / m)$ vs $\log$ P gives a straight line with positive slope i.e., $1 / n$

CHXII05:SURFACE CHEMISTRY

320821 The plot of $\frac{x}{m}$ Vs temperature at constant pressure is called

1 Adsorption isotherm

2 Adsorption isobar

3 Adsorption isochore

4 Freundlich isotherm

CHXII05:SURFACE CHEMISTRY

320818 The slope of the straight line graph between $log \frac{x}{m} and logP$ for the adsorption of a gas on solid is

1

k

2

\log k

3

n

4

\frac{1}{n}

Explanation:

$log(x/m) = (1/n)logP + logk y = mX + C$

CHXII05:SURFACE CHEMISTRY

320819 According to Freundlich adsorption isotherm, at high pressure, the value of $\frac{x}{m}$ is

1 Directly proportional to pressure

2 Inversely proportional to pressure

3 Directly proportional to square of pressure

4 Independent of pressure

Explanation:

$\frac{x}{m}$ doesn't depend on pressure.

CHXII05:SURFACE CHEMISTRY

320820 Which of the following graphs would yield a straight line?

1

\frac{x}{m} V s P

2

l o g (\frac{x}{m}) V s P

3

\frac{x}{m} V s l o g P

4

l o g (\frac{x}{m}) V s l o g P

Explanation:

According to Freundlich adsorption isotherm, $\log (x / m) = (1 / n) \log P + \log k$
$y = m x + C$
Plot of $\log (x / m)$ vs $\log$ P gives a straight line with positive slope i.e., $1 / n$

CHXII05:SURFACE CHEMISTRY

320821 The plot of $\frac{x}{m}$ Vs temperature at constant pressure is called

1 Adsorption isotherm

2 Adsorption isobar

3 Adsorption isochore

4 Freundlich isotherm

CHXII05:SURFACE CHEMISTRY

320818 The slope of the straight line graph between $log \frac{x}{m} and logP$ for the adsorption of a gas on solid is

1

k

2

\log k

3

n

4

\frac{1}{n}

Explanation:

$log(x/m) = (1/n)logP + logk y = mX + C$

CHXII05:SURFACE CHEMISTRY

320819 According to Freundlich adsorption isotherm, at high pressure, the value of $\frac{x}{m}$ is

1 Directly proportional to pressure

2 Inversely proportional to pressure

3 Directly proportional to square of pressure

4 Independent of pressure

Explanation:

$\frac{x}{m}$ doesn't depend on pressure.

CHXII05:SURFACE CHEMISTRY

320820 Which of the following graphs would yield a straight line?

1

\frac{x}{m} V s P

2

l o g (\frac{x}{m}) V s P

3

\frac{x}{m} V s l o g P

4

l o g (\frac{x}{m}) V s l o g P

Explanation:

According to Freundlich adsorption isotherm, $\log (x / m) = (1 / n) \log P + \log k$
$y = m x + C$
Plot of $\log (x / m)$ vs $\log$ P gives a straight line with positive slope i.e., $1 / n$

CHXII05:SURFACE CHEMISTRY

320821 The plot of $\frac{x}{m}$ Vs temperature at constant pressure is called

1 Adsorption isotherm

2 Adsorption isobar

3 Adsorption isochore

4 Freundlich isotherm

CHXII05:SURFACE CHEMISTRY

320818 The slope of the straight line graph between $log \frac{x}{m} and logP$ for the adsorption of a gas on solid is

1

k

2

\log k

3

n

4

\frac{1}{n}

Explanation:

$log(x/m) = (1/n)logP + logk y = mX + C$

CHXII05:SURFACE CHEMISTRY

320819 According to Freundlich adsorption isotherm, at high pressure, the value of $\frac{x}{m}$ is

1 Directly proportional to pressure

2 Inversely proportional to pressure

3 Directly proportional to square of pressure

4 Independent of pressure

Explanation:

$\frac{x}{m}$ doesn't depend on pressure.

CHXII05:SURFACE CHEMISTRY

320820 Which of the following graphs would yield a straight line?

1

\frac{x}{m} V s P

2

l o g (\frac{x}{m}) V s P

3

\frac{x}{m} V s l o g P

4

l o g (\frac{x}{m}) V s l o g P

Explanation:

According to Freundlich adsorption isotherm, $\log (x / m) = (1 / n) \log P + \log k$
$y = m x + C$
Plot of $\log (x / m)$ vs $\log$ P gives a straight line with positive slope i.e., $1 / n$

CHXII05:SURFACE CHEMISTRY

320821 The plot of $\frac{x}{m}$ Vs temperature at constant pressure is called

1 Adsorption isotherm

2 Adsorption isobar

3 Adsorption isochore

4 Freundlich isotherm