Question
Question: What is the approximate root means speed of line smoke particles of mass \({{10}^{-17}}kg\) at \({{2...
What is the approximate root means speed of line smoke particles of mass 10−17kg at 27oC?
(k=1.38×10−23Jk−1)
A. 2.0×10−2ms−1B. 2.7×10−2ms−1C. 3.5×10−2ms−1D. 4.4×10−2ms−1
Solution
Evaluate the kinetic energy in terms of molar mass m and then find kinetic energy in terms of temperature according to the kinetic theory of gases. After that, compare two relations of kinetic energy and find out the root mean speed.
Formula used: K.E=21mvrms2
K.E=23KT
Complete step-by-step solution
We know that the mass of a particle is 10−17kg and temperature27oCaccording to the kinetic theory of gases.
K.E=23KT......(1)
Where ‘k’ is Boltzmann constant having a value of 1.38×10−23 and ‘T’ is the temperature which is given in the question. We also know that,
K.E=21mvrms2.......(2)
Where ‘m’ is the mass of a particle and vrms is the root mean speed.
On comparing (1) and (2) equations of kinetic energy we get,
21mvrms2=23KT
After simplifying this equation, find the value of vrms that is the root mean speed.
vrms=m3KT
Now substitute the values of K,T and m in the equation, we get,
vrms=10−173×1.38×10−23×300
After calculating this we get the value of root mean speed.
vrms=3.5×10−2ms−1
Hence the correct option is (c).
Additional Information:
The square root of mean square speed is called root-mean-square speed or rms speed. It is denoted by the symbol vrms. And for a given sample of gas higher temperature means a higher value of vrms and similarly, lower temperature means a lower value of vrms.
Note: If any object having mass m and velocity of vrms then it is kinetic energy becomes 21mvrms2 because as we know that kinetic energy is the energy possessed by the motion of the object. Potential energy is another form of energy, which is possessed by an object by virtue of its position in a system.