Question

At \(T\)K, for one mole of an ideal gas, \(P\), \(V\) and \({v}_{rms}\) were measured. A graph of \(({v}_{rms})^2\) (on y-axis) and \(PV\) (on x-axis) goes straight line passing through origin. Its slope is \(m\). What is its molar mass (in kg mol\(^{-1}\))?

• A. \(3m\)
• B. \(\frac{1}{3m}\)
• C. \(\frac{m}{3}\)
• D. \(\frac{3}{m}\)

Hint:

Remember that the root-mean-square speed of a gas relates directly to the temperature and inversely to the molar mass.

Answer 1

The Correct Option is D

Step 1: Relate the given equation to the ideal gas law. \({PV} = {nRT}\), for one mole, \({PV} = {RT}\). \[ ({v}_{rms})^2 = {slope} \times {PV} = m \times {RT} \] Step 2: Use the relation for \({v}_{rms}\) in terms of molar mass. \[ {v}_{rms} = \sqrt{\frac{3RT}{M}} \] Equate and solve for \(M\): \[ \frac{3RT}{M} = m \times RT \rightarrow M = \frac{3}{m} \]