If the initial pressure of a gas is 0.03 atm, the mass of the gas adsorbed per gram of the adsorbent is____________× 10^-2g.

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Payal Gupta | Contributor-Level 10

3 months ago

Using, Freundlich adsorption isotherm ;

$\frac{x}{m} = k . p^{1 / n}$ ………………. (i)

$l o g \frac{x}{m} = \frac{1}{n} l o g p + l o g k$

Comparing with y = mx + C

Slope = $\frac{1}{n}$ = 1

Intercept, log k = 0.602

log k = log 4

k = 4

from equation (i)

$\frac{x}{m} = 4 \times {(0.03)}^{1}$

= 0.12

= 12 × 10^-²

So, 12 × 10^-² g of gas is adsorbed per gram of adsorbent,

Using, Freundlich adsorption isotherm ; <math> <mrow> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mi>m</mi> </mrow> </mfrac> <mo>=</mo> <mi>k</mi> <mo>.</mo> <msup> <mrow> <mi>p</mi> </mrow> <mrow> <mn>1</mn> <mo>/</mo> <mi>n</mi> </mrow> </msup> </mrow> </math>       ………………. (i) <math> <mrow> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mi>m</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>n</mi> </mrow> </mfrac> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mi>p</mi> <mo>+</mo> <mi>l</mi> <mi>o</mi> <mi>g</mi> <mi>k</mi> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1813568982"> </o:OLEObject></xml><! [endif]-> Comparing with y = mx + CSlope = <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>n</mi> </mrow> </mfrac> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1813568983"> </o:OLEObject></xml><! [endif]-> = 1Intercept, log k = 0.602log k = log 4k = 4from equation (i)  <math> <mrow> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mi>m</mi> </mrow> </mfrac> <mo>=</mo> <mn>4</mn> <mo>×</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>0</mn> <mo>.</mo> <mn>0</mn> <mn>3</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>1</mn> </mrow> </msup> </mrow> </math>             <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1813568984"> </o:OLEObject></xml><! [endif]-> = 0.12= 12 × 10-2So, 12 × 10-2 g of gas is adsorbed per gram of adsorbent,

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If the initial pressure of a gas is 0.03 atm, the mass of the gas adsorbed per gram of the adsorbent is____________× 10^-2g.

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If the initial pressure of a gas is 0.03 atm, the mass of the gas adsorbed per gram of the adsorbent is____________× 10^-2g.