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Answer in brief: Compare the rms speed of hydrogen molecules at 127°C with rms speed of oxygen molecules at 27ºC given that molecular masses of hydrogen and oxygen are 2 and 32 respectively. - Physics

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प्रश्न

Answer in brief:

Compare the rms speed of hydrogen molecules at 127°C with rms speed of oxygen molecules at 27ºC given that molecular masses of hydrogen and oxygen are 2 and 32 respectively.

Compare the r.m.s. speed of hydrogen molecule at 127°C with r.m.s. speed of oxygen molecule at 27°C, given that molecular masses of hydrogen and oxygen are 2 and 32 respectively.

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उत्तर

Data: M01 (hydrogen) = 2 g/mol

M02 (oxygen) = 32 g/mol

T1 (hydrogen) = 273 + 127 = 400 K

T2 (oxygen) = 273 + 27 = 300 K

The r.m.s. speed, `(v_(r.m.s.)) = sqrt((3 R T)/M_0)`,

Where M0 denotes the molar mass.

∴ `(v_(r.m.s._1) ("hydrogen"))/(v_(r.m.s._2) ("oxygen")) = sqrt((T_1/T_2)(M_02/M_01))`

= `sqrt((400/300)(32/2))`

= `sqrt((4/3)(16))`

= `((2)(4))/sqrt3`

= `8/sqrt3`

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Interpretation of Temperature in Kinetic Theory
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अध्याय 3: Kinetic Theory of Gases and Radiation - Exercises [पृष्ठ ७४]

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बालभारती Physics [English] Standard 12 Maharashtra State Board
अध्याय 3 Kinetic Theory of Gases and Radiation
Exercises | Q 16 | पृष्ठ ७४

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