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Grade upto college level Physical Chemistry

A gas bulb of 1 litre capacity contains 2.0 * 1021 molecules of nitrogen exerting a pressure of 7.57 * 103 Nm-2. Calculate the root mean square (r.m.s) speed and the temperature of the gas molecules. If the ratio of the most probable speed to the root mean square speed is 0.82, calculate the most probable speed for these molecules at this temperature.

Profile image of Amit Saxena
12 Years agoGrade upto college level
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1 Answer

Profile image of Navjyot Kalra
12 Years ago
Vrms = √3RT/M = √3P/d …(1)
Calculation of density (d) of gas molecules.
No. of gas molecules = 2 * 1021
Since mass of 6.023 * 1023 molecules of nitrogen = 28 g
Mass of 2 * 1021 molecules of N2
= 28 * 2 *1021/6.023 * 1023 = 56/602.3 = 0.093 g
∴ d = Mass/Volume = 0.093/1 g/1 [volume of flask = 1 L]
= 0.093 * 10-3 kg/10-3 m3 = 0.093 kg/m3
Substituting the value of d and P in equation (1)
Vrms = √3 * 7.57 *103/0.093 = 494.16 m/sec [Vrms = √3P/d]
(Vrms)2 = 3 RT/M = 3P/d = 3 * 7.57 * 103/0.093 [squaring]
RT/M 7.57 * 103/0.093
T = 7.57 * 103/0.093 * 28 * 10-3/8.314 = 274.13 K
Most probable velocity (∝)/Root mean sq. velocity (Vrms) = 0.82 [given]
=> ∝ = 0.82 * 494.16 = 405.2 m/sec
ALTERNATESOLUTION
Given V = 1L = 10-3m3, P = 7.57 * 10-3 Nm-2, R = 8.314J,
N = 2 * 1021/6.023 * 1023 moles
PV = nRT
or T = PV/nR
= 7.57 *10-3 * 6.023 *1023/2 *1021 * 8.31 = 274.13 K
Urms = √3RT/M =√3 * 8.314 * 274.13/28 *10-3 m/s = 494.15 m/s
(Given U)
Ump/Urms = 0.82 (given)
∴ Ump = 0.82 * Urms = 0.82 * 494.15 = 405.2 m/sec

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