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2. The rms speed of molecules in a gas at `20.0°C` is to be increased by `1.0%`. To what temperature must it be raised?
2. The rms speed of molecules in a gas at 20.0°C is to be increased by 1.0%. To what temperature must it be raised? `v_(rms) = sqrt((3kT)/m)`
`299.1\ K (25.9°C)`
Given:
`T_1 = 20°C = 293\ K`
`v_2 = 1.010v_1`
Equation:
`v_(rms) = sqrt((3kT)/m)`
Solution:
`v_2 = 1.010v_1`
`=>`
`sqrt((3kT_2)/m) = 1.010sqrt((3kT_1)/m)`
`=>`
`sqrt((3kT_2)/m)/sqrt((3kT_1)/m) = 1.010`
`=>`
`sqrt(T_2/T_1) = 1.010`
`T_2=T_1(1.010)^2`
`T_2=(293\ K)(1.010)^2`
`T_2= 299.1\ K (25.9°C)` |