320813
Statement A : A gas with higher critical temperature is adsorbed more than a gas with lower critical temperature. Statement B : Higher critical temperature means that the gas is more easily liquefiable.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
More easily liquefiable gases are easily adsorbed and a gas having higher critical temperature is more easily liquefiable. So, the option (3) is correct.
CHXII05:SURFACE CHEMISTRY
320814
In Freundlich adsorption isotherm, the value of \({\rm{1/n}}\) is
1 Between 0 and 1 in all cases
2 Between 2 and 4 in all cases
3 1 in case of physical adsorption
4 1 in case of chemisorption
Explanation:
Freundlich adsorption isotherm is \(\frac{{\rm{x}}}{{\rm{m}}}{\rm{ = k}}{{\rm{P}}^{{\rm{1/n}}}}\) \({\text{If}}\,\,{\text{P}} \to {\text{0,n = 1}}\;\;\;\therefore \frac{{\text{x}}}{{\text{m}}}{\text{ = kP}}\) \({\text{If P is high;n = 0}}\;\;\therefore \;\;\frac{{\text{x}}}{{\text{m}}}{\text{ = k}}{{\text{P}}^{\text{0}}}\)
CHXII05:SURFACE CHEMISTRY
320815
Based on Langmuir adsorption isotherm, the intercept in the graph \(\left( {\frac{{\rm{m}}}{{\rm{x}}}} \right.{\rm{versus}}\,\,\left. {\frac{{\rm{1}}}{{\rm{p}}}} \right)\) is equal to
320816
For a linear plot of \(\log (\mathrm{x} / \mathrm{m})\) versus \(\log \mathrm{p}\) in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and \(\mathrm{n}\) are constants)
1 \({\text{1 / n}}\) appears as the intercept
2 Only \({\text{1 / n}}\) appears as the slope
3 \({\rm{log(1/n)}}\) appears as the intercept
4 Both \(\mathrm{k}\) and \(1 / \mathrm{n}\) appear in the slope term
320817
The volume of gases \(\mathrm{H}_{2}, \mathrm{CH}_{4}, \mathrm{CO}_{2}\) and \(\mathrm{NH}_{3}\) adsorbed by \(1 \mathrm{~g}\) of charcoal at \(288 \mathrm{~K}\) are in the order of
A gas having highest critical temperature will be adsorbed on a solid to a greater extent The order of critical temperature is \({\text{N}}{{\text{H}}_{\text{3}}}{\text{ > C}}{{\text{O}}_{\text{2}}}{\text{ > C}}{{\text{H}}_{\text{4}}}{\text{ > }}{{\text{H}}_{\text{2}}}\) Hence, the volume of gas adsorbed in the order of \({\text{N}}{{\text{H}}_{\text{3}}}{\text{ > C}}{{\text{O}}_{\text{2}}}{\text{ > C}}{{\text{H}}_{\text{4}}}{\text{ > }}{{\text{H}}_{\text{2}}}\)
320813
Statement A : A gas with higher critical temperature is adsorbed more than a gas with lower critical temperature. Statement B : Higher critical temperature means that the gas is more easily liquefiable.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
More easily liquefiable gases are easily adsorbed and a gas having higher critical temperature is more easily liquefiable. So, the option (3) is correct.
CHXII05:SURFACE CHEMISTRY
320814
In Freundlich adsorption isotherm, the value of \({\rm{1/n}}\) is
1 Between 0 and 1 in all cases
2 Between 2 and 4 in all cases
3 1 in case of physical adsorption
4 1 in case of chemisorption
Explanation:
Freundlich adsorption isotherm is \(\frac{{\rm{x}}}{{\rm{m}}}{\rm{ = k}}{{\rm{P}}^{{\rm{1/n}}}}\) \({\text{If}}\,\,{\text{P}} \to {\text{0,n = 1}}\;\;\;\therefore \frac{{\text{x}}}{{\text{m}}}{\text{ = kP}}\) \({\text{If P is high;n = 0}}\;\;\therefore \;\;\frac{{\text{x}}}{{\text{m}}}{\text{ = k}}{{\text{P}}^{\text{0}}}\)
CHXII05:SURFACE CHEMISTRY
320815
Based on Langmuir adsorption isotherm, the intercept in the graph \(\left( {\frac{{\rm{m}}}{{\rm{x}}}} \right.{\rm{versus}}\,\,\left. {\frac{{\rm{1}}}{{\rm{p}}}} \right)\) is equal to
320816
For a linear plot of \(\log (\mathrm{x} / \mathrm{m})\) versus \(\log \mathrm{p}\) in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and \(\mathrm{n}\) are constants)
1 \({\text{1 / n}}\) appears as the intercept
2 Only \({\text{1 / n}}\) appears as the slope
3 \({\rm{log(1/n)}}\) appears as the intercept
4 Both \(\mathrm{k}\) and \(1 / \mathrm{n}\) appear in the slope term
320817
The volume of gases \(\mathrm{H}_{2}, \mathrm{CH}_{4}, \mathrm{CO}_{2}\) and \(\mathrm{NH}_{3}\) adsorbed by \(1 \mathrm{~g}\) of charcoal at \(288 \mathrm{~K}\) are in the order of
A gas having highest critical temperature will be adsorbed on a solid to a greater extent The order of critical temperature is \({\text{N}}{{\text{H}}_{\text{3}}}{\text{ > C}}{{\text{O}}_{\text{2}}}{\text{ > C}}{{\text{H}}_{\text{4}}}{\text{ > }}{{\text{H}}_{\text{2}}}\) Hence, the volume of gas adsorbed in the order of \({\text{N}}{{\text{H}}_{\text{3}}}{\text{ > C}}{{\text{O}}_{\text{2}}}{\text{ > C}}{{\text{H}}_{\text{4}}}{\text{ > }}{{\text{H}}_{\text{2}}}\)
320813
Statement A : A gas with higher critical temperature is adsorbed more than a gas with lower critical temperature. Statement B : Higher critical temperature means that the gas is more easily liquefiable.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
More easily liquefiable gases are easily adsorbed and a gas having higher critical temperature is more easily liquefiable. So, the option (3) is correct.
CHXII05:SURFACE CHEMISTRY
320814
In Freundlich adsorption isotherm, the value of \({\rm{1/n}}\) is
1 Between 0 and 1 in all cases
2 Between 2 and 4 in all cases
3 1 in case of physical adsorption
4 1 in case of chemisorption
Explanation:
Freundlich adsorption isotherm is \(\frac{{\rm{x}}}{{\rm{m}}}{\rm{ = k}}{{\rm{P}}^{{\rm{1/n}}}}\) \({\text{If}}\,\,{\text{P}} \to {\text{0,n = 1}}\;\;\;\therefore \frac{{\text{x}}}{{\text{m}}}{\text{ = kP}}\) \({\text{If P is high;n = 0}}\;\;\therefore \;\;\frac{{\text{x}}}{{\text{m}}}{\text{ = k}}{{\text{P}}^{\text{0}}}\)
CHXII05:SURFACE CHEMISTRY
320815
Based on Langmuir adsorption isotherm, the intercept in the graph \(\left( {\frac{{\rm{m}}}{{\rm{x}}}} \right.{\rm{versus}}\,\,\left. {\frac{{\rm{1}}}{{\rm{p}}}} \right)\) is equal to
320816
For a linear plot of \(\log (\mathrm{x} / \mathrm{m})\) versus \(\log \mathrm{p}\) in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and \(\mathrm{n}\) are constants)
1 \({\text{1 / n}}\) appears as the intercept
2 Only \({\text{1 / n}}\) appears as the slope
3 \({\rm{log(1/n)}}\) appears as the intercept
4 Both \(\mathrm{k}\) and \(1 / \mathrm{n}\) appear in the slope term
320817
The volume of gases \(\mathrm{H}_{2}, \mathrm{CH}_{4}, \mathrm{CO}_{2}\) and \(\mathrm{NH}_{3}\) adsorbed by \(1 \mathrm{~g}\) of charcoal at \(288 \mathrm{~K}\) are in the order of
A gas having highest critical temperature will be adsorbed on a solid to a greater extent The order of critical temperature is \({\text{N}}{{\text{H}}_{\text{3}}}{\text{ > C}}{{\text{O}}_{\text{2}}}{\text{ > C}}{{\text{H}}_{\text{4}}}{\text{ > }}{{\text{H}}_{\text{2}}}\) Hence, the volume of gas adsorbed in the order of \({\text{N}}{{\text{H}}_{\text{3}}}{\text{ > C}}{{\text{O}}_{\text{2}}}{\text{ > C}}{{\text{H}}_{\text{4}}}{\text{ > }}{{\text{H}}_{\text{2}}}\)
320813
Statement A : A gas with higher critical temperature is adsorbed more than a gas with lower critical temperature. Statement B : Higher critical temperature means that the gas is more easily liquefiable.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
More easily liquefiable gases are easily adsorbed and a gas having higher critical temperature is more easily liquefiable. So, the option (3) is correct.
CHXII05:SURFACE CHEMISTRY
320814
In Freundlich adsorption isotherm, the value of \({\rm{1/n}}\) is
1 Between 0 and 1 in all cases
2 Between 2 and 4 in all cases
3 1 in case of physical adsorption
4 1 in case of chemisorption
Explanation:
Freundlich adsorption isotherm is \(\frac{{\rm{x}}}{{\rm{m}}}{\rm{ = k}}{{\rm{P}}^{{\rm{1/n}}}}\) \({\text{If}}\,\,{\text{P}} \to {\text{0,n = 1}}\;\;\;\therefore \frac{{\text{x}}}{{\text{m}}}{\text{ = kP}}\) \({\text{If P is high;n = 0}}\;\;\therefore \;\;\frac{{\text{x}}}{{\text{m}}}{\text{ = k}}{{\text{P}}^{\text{0}}}\)
CHXII05:SURFACE CHEMISTRY
320815
Based on Langmuir adsorption isotherm, the intercept in the graph \(\left( {\frac{{\rm{m}}}{{\rm{x}}}} \right.{\rm{versus}}\,\,\left. {\frac{{\rm{1}}}{{\rm{p}}}} \right)\) is equal to
320816
For a linear plot of \(\log (\mathrm{x} / \mathrm{m})\) versus \(\log \mathrm{p}\) in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and \(\mathrm{n}\) are constants)
1 \({\text{1 / n}}\) appears as the intercept
2 Only \({\text{1 / n}}\) appears as the slope
3 \({\rm{log(1/n)}}\) appears as the intercept
4 Both \(\mathrm{k}\) and \(1 / \mathrm{n}\) appear in the slope term
320817
The volume of gases \(\mathrm{H}_{2}, \mathrm{CH}_{4}, \mathrm{CO}_{2}\) and \(\mathrm{NH}_{3}\) adsorbed by \(1 \mathrm{~g}\) of charcoal at \(288 \mathrm{~K}\) are in the order of
A gas having highest critical temperature will be adsorbed on a solid to a greater extent The order of critical temperature is \({\text{N}}{{\text{H}}_{\text{3}}}{\text{ > C}}{{\text{O}}_{\text{2}}}{\text{ > C}}{{\text{H}}_{\text{4}}}{\text{ > }}{{\text{H}}_{\text{2}}}\) Hence, the volume of gas adsorbed in the order of \({\text{N}}{{\text{H}}_{\text{3}}}{\text{ > C}}{{\text{O}}_{\text{2}}}{\text{ > C}}{{\text{H}}_{\text{4}}}{\text{ > }}{{\text{H}}_{\text{2}}}\)
320813
Statement A : A gas with higher critical temperature is adsorbed more than a gas with lower critical temperature. Statement B : Higher critical temperature means that the gas is more easily liquefiable.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
More easily liquefiable gases are easily adsorbed and a gas having higher critical temperature is more easily liquefiable. So, the option (3) is correct.
CHXII05:SURFACE CHEMISTRY
320814
In Freundlich adsorption isotherm, the value of \({\rm{1/n}}\) is
1 Between 0 and 1 in all cases
2 Between 2 and 4 in all cases
3 1 in case of physical adsorption
4 1 in case of chemisorption
Explanation:
Freundlich adsorption isotherm is \(\frac{{\rm{x}}}{{\rm{m}}}{\rm{ = k}}{{\rm{P}}^{{\rm{1/n}}}}\) \({\text{If}}\,\,{\text{P}} \to {\text{0,n = 1}}\;\;\;\therefore \frac{{\text{x}}}{{\text{m}}}{\text{ = kP}}\) \({\text{If P is high;n = 0}}\;\;\therefore \;\;\frac{{\text{x}}}{{\text{m}}}{\text{ = k}}{{\text{P}}^{\text{0}}}\)
CHXII05:SURFACE CHEMISTRY
320815
Based on Langmuir adsorption isotherm, the intercept in the graph \(\left( {\frac{{\rm{m}}}{{\rm{x}}}} \right.{\rm{versus}}\,\,\left. {\frac{{\rm{1}}}{{\rm{p}}}} \right)\) is equal to
320816
For a linear plot of \(\log (\mathrm{x} / \mathrm{m})\) versus \(\log \mathrm{p}\) in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and \(\mathrm{n}\) are constants)
1 \({\text{1 / n}}\) appears as the intercept
2 Only \({\text{1 / n}}\) appears as the slope
3 \({\rm{log(1/n)}}\) appears as the intercept
4 Both \(\mathrm{k}\) and \(1 / \mathrm{n}\) appear in the slope term
320817
The volume of gases \(\mathrm{H}_{2}, \mathrm{CH}_{4}, \mathrm{CO}_{2}\) and \(\mathrm{NH}_{3}\) adsorbed by \(1 \mathrm{~g}\) of charcoal at \(288 \mathrm{~K}\) are in the order of
A gas having highest critical temperature will be adsorbed on a solid to a greater extent The order of critical temperature is \({\text{N}}{{\text{H}}_{\text{3}}}{\text{ > C}}{{\text{O}}_{\text{2}}}{\text{ > C}}{{\text{H}}_{\text{4}}}{\text{ > }}{{\text{H}}_{\text{2}}}\) Hence, the volume of gas adsorbed in the order of \({\text{N}}{{\text{H}}_{\text{3}}}{\text{ > C}}{{\text{O}}_{\text{2}}}{\text{ > C}}{{\text{H}}_{\text{4}}}{\text{ > }}{{\text{H}}_{\text{2}}}\)
