For one mole of an ideal gas, the slope of $V$ vs. $T$ curve at constant pressure of 2 atm is $X$ L…
For one mole of an ideal gas, the slope of $V$ vs. $T$ curve at constant pressure of 2 atm is $X$ L $\mathrm{mol}^{-1} \mathrm{K}^{-1} .$ The value of the ideal universal gas constant 'R' in terms of $X$ is
$X$ L atm mol $^{-1} \mathrm{K}^{-1}$
$\frac{x}{2}$ L atm $\mathrm{mol}^{-1} \mathrm{K}^{-1}$
At constant pressure and same number of moles, Volume of the gas is the directly proportional to the temperature of the gas. The slope of V versus T graph i.e slope \(=\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\mathrm{X} \mathrm{Lmol}^{-1} \mathrm{~K}^{-1}\)
But, \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{T}}=\frac{\mathrm{R}}{\mathrm{P}}\) (no. of moles \(=1\))
Now, \(\mathrm{P}=2\)
Therefore, \(\mathrm{R}=2 \mathrm{~X~} \mathrm{L~} \mathrm{atm~} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\)
,