At $T(K)$ for one mole of an ideal gas, the graph of $P$ (on $y$-axis) and $\mathrm{V}^{-1}$ (on $x$-axis)…

At $T(K)$ for one mole of an ideal gas, the graph of $P$ (on $y$-axis) and $\mathrm{V}^{-1}$ (on $x$-axis) gave a straight line with slope of $32.8 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1}$. What is the temperature (in K )? $\left(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm}^{-1} \mathrm{Kol}^{-1}\right)$
  1. 600
  2. 200
  3. 800
  4. 400

Solution

Ideal gas equation is $\begin{aligned} & \mathrm{PV}=\mathrm{nRT} \\ & \text { for } 1 \mathrm{~mole}, \\ & \mathrm{P}=\frac{1}{\mathrm{~V}} \mathrm{RT} \\ & \mathrm{Y}=\mathrm{x} \mathrm{m}+\mathrm{C} \\ & \mathrm{m} \text { (slope) }=\mathrm{RT} \\ & \mathrm{C} \text { (intercept) }=0 \\ & \text { Slope }(\mathrm{m})=32.8 \mathrm{Latm} \mathrm{mol}^{-1} \\ & \therefore \quad \mathrm{RT}=32.8 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \\ & \text { or, } \mathrm{T}=\frac{32.8 \mathrm{Latmmol}^{-1}}{0.082 \mathrm{atmLmol}}{ }^{-1} \mathrm{~K}^{-1} \end{aligned}$ $\mathrm{T}=400 \mathrm{~K}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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