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In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:
At constant temperature and constant pressure for a fixed mass of gaseous mixtures,
PV = nRT Volume ∝ number of moles
Where Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200°C and 252°C are 70.2 and 57.2 respectively at one atmosphere
Q.
From the given data, it can be interpreted that .
  • a)
     If degree of dissociation is more, then vapour density is more for PCI5
  • b)
    More the degree of dissociation, less is the vapour density of PCI5
  • c)
    Degree of dissociation and vapour density cannot be correlated.
  • d)
    a is independent of temperature.
Correct answer is option 'B'. Can you explain this answer?
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In gaseous dissociation reactions, the total mass remains unchanged an...
More the degree of dissociation, less is the vapour density of PCI5.
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In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.The observation shows that the dissociation constants (KP) for the reaction are

In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows that

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculatedfrom their molecular formula. The observed densities decrease towards a limit as thetemperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.Examples :With increase in the number of molecules, the volume increases (pressure remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,t = 0 at = teq a x nx Total no. of moles Observed molecular weight or molar mass of the mixtureQ.The K for the reaction is 640 mm at 775 K. The percentage dissociation of N2O4 at equilibrium pressure of 160 mm is

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculatedfrom their molecular formula. The observed densities decrease towards a limit as thetemperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.Examples :With increase in the number of molecules, the volume increases (pressure remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,t = 0 at = teq a x nx Total no. of moles Observed molecular weight or molar mass of the mixtureQ.A sample of mixture of A(g), B(g) and C(g) under equilibrium has a mean molecular weight (observe d) is 80.The equilibrium is Find the degree of dissociation for A(g).

The atmospheric lapse rateFor small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantityTP(1 – λ)/λ has a uniform value through the layers of troposphere.(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)Q. Mechanical equilibrium of the atmosphere requires that the pressure decreases with altitude according to . Assuming free fall acceleration to be uniform, then lapse rate is given by

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In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer?
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In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer?.
Solutions for In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer?, a detailed solution for In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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