Question

In the mixture 2 mole of $\mathrm{He}$ and 1 mole of $\mathrm{Ar}$ is present. Find $[{\left({\mathrm{V}}_{\mathrm{RMS}}\right)}_{\mathrm{He}}/{\left({\mathrm{V}}_{\mathrm{RMS}}\right)}_{\mathrm{Ar}}]$at 300 K.

• A. 6.32
• B. 1.58
• C. 3.16
• D. 10

Answer 1

The correct option is C

3.16

Explanation of correct option:

Option C is correct.

${\mathrm{V}}_{\mathrm{RMS}}$ is the root mean square voltage of the sinusoidal source of electromotive force. It is used to find out the value of AC supply.

$V_{RMS}=\sqrt{\frac{3RT}{M}}$thus, the ratio for both helium and argon is:

$\frac{{\left(V_{RMS}\right)}_{He}}{{\left(V_{RMS}\right)}_{Ar}}=\sqrt{\frac{{\displaystyle \raisebox{1ex}{$3RT$}\!\left/ \!\raisebox{-1ex}{$M_{He}$}\right.}}{\raisebox{1ex}{$3RT$}\!\left/ \!\raisebox{-1ex}{$M_{Ar}$}\right.}}$

$\frac{{\left(V_{RMS}\right)}_{He}}{{\left(V_{RMS}\right)}_{Ar}}=\sqrt{\frac{40}{4}}$

$\frac{{\left(V_{RMS}\right)}_{He}}{{\left(V_{RMS}\right)}_{Ar}}=3.16$

Thus, the correct option is C.

Final Answer: Hence, the correct option is C.