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Uranium has two isotopes of masses 235 a...

Uranium has two isotopes of masses 235 and 238 units. If both are present in uranium hexa fluoride gas, which would have the larger average speed ? If atomic mass of fluorine is 19 units, estimate the percentage difference in speed at any temperature.

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To solve the problem, we need to determine which isotope of uranium (U-235 or U-238) has a larger average speed when present in uranium hexafluoride (UF6) and estimate the percentage difference in their speeds at any temperature. ### Step-by-Step Solution: 1. **Identify the Masses:** - The mass of U-235 = 235 units - The mass of U-238 = 238 units - The atomic mass of fluorine (F) = 19 units ...
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Uranium. has two isotopes of masses 235 and 238 units. If both of them are present in Uranium hexafluoride gas, find the percentage ratio of difference in rms velocities of two isotopes to the rms velocity of heavier isotope.

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The speed of a molecule of a gas changes continuously as a result of collisions with other molecules and with the walls of the container. The speeds of individual molecules therefore change with time. A direct consequence of the distribution of speeds is that the average kinetric energy is constant for a given temperature. The average K.E. is defined as bar(KE) =1/N((1)/(2)m underset(i)SigmadN_(i)u_(1)^(2)) = 1/2 m(underset(i)Sigma(dN_(i))/(N).u_(1)^(2)) where (dN)/(N) is the fraction of molecules having speeds between u_(i) and u_(i) + du and as proposed by maxwell (dN)/(N) = 4pi ((m)/(2pi KT))^(3//2) exp (-("mu"^(2))/(2KT)).u^(2).du The plot of ((1)/(N) (dN)/(du)) is plotted for a particular gas at two different temperature against u as shown. The majority of molecules have speeds which cluster around "v"_("MPS") in the middle of the range of v. There area under the curve between any two speeds V_(1) "and" V_(2) is the fraction of molecules having speeds between V_(1) "and" V_(2). The speed distribution also depends on the mass of the molecules. As the area under the curve is the same (equal to unity) for all gas samples, samples which have the same "V"_("MPS") will have identical Maxwellian plots. On the basis of the above passage answer the questions that follow. For the above graph, drawn for two different samples of gases at two different temperature T_(1) "and" T_(2). Which of the following statements is necessarily true?

The speed of a molecule of a gas changes continuously as a result of collisions with other molecules and with the walls of the container. The speeds of individual molecules therefore change with time. A direct consequence of the distribution of speeds is that the average kinetric energy is constant for a given temperature. The average K.E. is defined as bar(KE) =1/N((1)/(2)m underset(i)SigmadN_(i)u_(1)^(2)) = 1/2 m(underset(i)Sigma(dN_(i))/(N).u_(1)^(2)) where (dN)/(N) is the fraction of molecules having speeds between u_(i) and u_(i) + du and as proposed by maxwell (dN)/(N) = 4pi ((m)/(2pi KT))^(3//2) exp (-("mu"^(2))/(2KT)).u^(2).du The plot of ((1)/(N) (dN)/(du)) is plotted for a particular gas at two different temperature against u as shown. The majority of molecules have speeds which cluster around "v"_("MPS") in the middle of the range of v. There area under the curve between any two speeds V_(1) "and" V_(2) is the fraction of molecules having speeds between V_(1) "and" V_(2). The speed distribution also depends on the mass of the molecules. As the area under the curve is the same (equal to unity) for all gas samples, samples which have the same "V"_("MPS") will have identical Maxwellian plots. On the basis of the above passage answer the questions that follow. If two gases A and B and at temperature T_(A) "and"T_(B) respectivley have identical Maxwellian plots then which of the following statement are true?

The speed of a molecule of a gas changes continuously as a result of collisions with other molecules and with the walls of the container. The speeds of individual molecules therefore change with time. A direct consequence of the distribution of speeds is that the average kinetric energy is constant for a given temperature. The average K.E. is defined as bar(KE) =1/N((1)/(2)m underset(i)SigmadN_(i)u_(1)^(2)) = 1/2 m(underset(i)Sigma(dN_(i))/(N).u_(1)^(2)) where (dN)/(N) is the fraction of molecules having speeds between u_(i) and u_(i) + du and as proposed by maxwell (dN)/(N) = 4pi ((m)/(2pi KT))^(3//2) exp (-("mu"^(2))/(2KT)).u^(2).du The plot of ((1)/(N) (dN)/(du)) is plotted for a particular gas at two different temperature against u as shown. The majority of molecules have speeds which cluster around "v"_("MPS") in the middle of the range of v. There area under the curve between any two speeds V_(1) "and" V_(2) is the fraction of molecules having speeds between V_(1) "and" V_(2). The speed distribution also depends on the mass of the molecules. As the area under the curve is the same (equal to unity) for all gas samples, samples which have the same "V"_("MPS") will have identical Maxwellian plots. On the basis of the above passage answer the questions that follow. If a gas sample contains a total of N molecules. The area under any given maxwellian plot is equal ot :

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