A vessel is filled with 0.014 kg0.014 kg Nitrogen gas at a temperature of 27°C27°C. Let the νrmsνrms at this temperature be VV; then at what temperature the νrmsνrms of this gas will be 2V2V
Question
A vessel is filled with 0.014 kg0.014 kg Nitrogen gas at a temperature of 27°C27°C. Let the νrmsνrms at this temperature be VV; then at what temperature the νrmsνrms of this gas will be 2V2V
Solution
The root mean square speed (νrms) of a gas is given by the formula:
νrms = sqrt(3kT/m)
where: k is the Boltzmann constant, T is the temperature in Kelvin, m is the mass of one molecule of the gas.
Given that the root mean square speed at 27°C (or 300.15K) is V, we can write:
V = sqrt(3kT/m)
We want to find the temperature at which the root mean square speed is 2V. We can write this as:
2V = sqrt(3kT'/m)
where T' is the new temperature we want to find.
Squaring both sides of the equation gives:
4V^2 = 3kT'/m
Substituting the first equation into this gives:
4(3kT/m) = 3kT'/m
Solving for T' gives:
T' = 4T
So the temperature at which the root mean square speed is twice the original speed is four times the original temperature. Therefore, the new temperature is 4 * 300.15K = 1200.6K.
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