RMS velocity of one mole of an ideal gas was measured at different temperatures. A graph of…

RMS velocity of one mole of an ideal gas was measured at different temperatures. A graph of $\left(\mathrm{u}_{\mathrm{rms}}\right)^2$ (on y -axis) and $\mathrm{T}(\mathrm{K})$ (on x -axis) gave straight line passing through the origin and its slope is $249 \mathrm{~m}^2 \mathrm{~s}^{-2} \mathrm{~K}^{-1}$. What is the molar mass (in $\left.\mathrm{kg} \mathrm{mol}^{-1}\right)$ of ideal gas? $\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$
  1. $10$
  2. $1.0$
  3. $24.9$
  4. $1 \times 10^{-1}$

Solution

RMS velocity $\left(v_{\text {rms }}\right)=\sqrt{\frac{3 R T}{M}}$ $\begin{aligned} & v_{\mathrm{rms}}^2=\frac{3 \mathrm{RT}}{\mathrm{M}} \\ & \mathrm{y}=\mathrm{mx}+\mathrm{C}\end{aligned}$

where, $\begin{aligned}& \mathrm{y}=\mathrm{v}_{\mathrm{rms}}^2 \\ & \mathrm{x}=\mathrm{T}\end{aligned}$ $\begin{aligned} & \mathrm{m}(\text { slope })=\frac{3 R}{M} \\ & C(\text { intercept })=0\end{aligned}$ Slope $(\mathrm{m})=\frac{3 \mathrm{R}}{\mathrm{M}} \quad\left[\mathrm{R}=83 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right]$ $\frac{3 R}{M}=249 \mathrm{~m}^2 \mathrm{~s}^{-2} \mathrm{~K}^{-1}$ $\mathrm{M}=\frac{3 \times 8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}}{249 \mathrm{~m}^2 \mathrm{~s}^{-2} \mathrm{~K}^{-1}} \quad[1 \mathrm{~J}=1 \mathrm{~kg}]$ $\mathrm{M}=\frac{3 \times 8.3 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}}{249 \mathrm{~m}^2 \mathrm{~s}^2 \mathrm{~K}^{-1}}\left[1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}\right]$ $\begin{aligned} & \mathrm{M}=\frac{24.9 \mathrm{~kg} \mathrm{~mol}^{-1}}{249} \\ & \mathrm{M}=0.1 \mathrm{~kg} \mathrm{~mol}^{-1} \\ & \therefore \mathrm{M} \text { (molar mass) }=1 \times 10^{-1} \mathrm{~kg} \mathrm{~mol}^{-1}\end{aligned}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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