Mechanical Engineering - Thermodynamics - Discussion

Formula for most probable velocity.

vp = option B.

And for root mean square velocity.

Vrms=option C.

k = The molar gas constant has the value 8.3143 J/mol K.

T = Temperature.

Answer B is right.

Most probable velocity vp = root of 2kT/m.

The most probable speed, vp, is the speed most likely to be possessed by any molecule (of the same mass m) in the system and corresponds to the maximum value or mode of f(v). To find it, we calculate df/dv, set it to zero and solve for v:

\frac{df(v)}{dv} = 0

which yields:

v_p = \sqrt {\frac{2kT}{m}} = \sqrt {\frac{2RT}{M}}


The root mean square speed is the second-order moment of speed:

= \sqrt {\langle v^2 \rangle} = \left(\int_0^{\infty} v^2 \, f(v) \, dv \right)^{1/2}.

= \sqrt {\frac{3kT}{m}} = \sqrt {\frac{3RT}{M} } = \sqrt{\frac{3}{2}} v_p.

Which book is good for thermodynamics? Can anyone tell me?

@Venkatarao, P K Nag and Cengel.

I think answer is option C.

μ rms = √ (3RT/M).

where
μrms = root mean square velocity in m/sec.
R = ideal gas constant = 8.3145 (kg.m2/sec2)/Kmol.
T = absolute temperature in Kelvin.
M = mass of a mole of the gas in kilograms.

Then how to find most probable velocity?

Please mention the method just like you explained about root mean square velocity.

I want the correct explanation for this question. Please someone describe it clearly.