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# Scan_Doc0119 - What is the mass of a mole of He in...

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R = 0.08206 L . atm K· mol J R = 8.3145- K -- 1 . mo EXAMPLE 5.19 5.6 The Kinetic Molecular Theory of Gases 211 Before wecan use this equation, we need to consider the units for R. So far we have used 0.08206 L . atm/K . mol as the value of R. But to obtain the desired units (meters per second) for u rms , R must be expressed in different units. As we will see in more de- tail in Chapter 6, the energy unit most often used in the SI system is the joule (J). A joule is defined as a kilogram meter squared per second squared (kg' m2/s2).When R is con- verted to include the unit of joules, it has the value 8.3145 J/K . mol. When R in these units is used in the expression V3RT/M, U rms is obtained in the units of meters per second as desired. Calculate the root mean square velocity for the atoms in a sample of helium gas at 25°C. Solution Where are we going? • To determine the root mean square velocity for the atoms of He What do we know? ./ T = 25°C + 273 = 298 K ./ R = 8.3145 J/K . mol What information do we need? ./ Root mean square velocity is
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Unformatted text preview: What is the mass of a mole of He in kilograms? g 1ks M = 4.00-X --"'-= 4.00 X 10-3 kg/mol mol 1000 g What is the root mean square velocity for the atoms of He? 3(8.3145 J )(298 R) ~ R·mcl J k = 1.86 X 1Q6 k 400 X 1O-3-.L g . mol Since the units of J are kg . m2/s2, this expression gives ~ kg. m2 • U rms = 1.86 X 10 6 2 = 1.36 X 10 3 m/s kg. s Reality Check: The resulting units are appropriate for velocity. SEE EXERCISES 5.97 AND 5.98 So far we have said nothing about the range of velocities actually found in a gas sam-ple. In a real gas there are large numbers of collisions between particles. For example, as we will see in the next section, when an odorous gas such as ammonia is released in a room, it takes some time for the odor to permeate the air. This delay results from colli-sions between the NH3 molecules and the O 2 and N2 molecules in the air, which greatly slow the mixing process....
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