At T (K),the P,V and u_{rms} of 1 mole of an | vedclass

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(A) The root mean square velocity $(u_{rms})$ is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.For $1$ mole of an ideal gas,the ideal gas equ

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$\frac{3}{M}$

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(A) The root mean square velocity $(u_{rms})$ is given by the formula: $u_{rms} = \sqrt{\frac{3RT}{M}}$.For $1$ mole of an ideal gas,the ideal gas equation is $PV = RT$.Substituting $RT = PV$ into the $u_{rms}$ equation,we get:$u_{rms} = \sqrt{\frac{3PV}{M}}$Squaring both sides,we get:$u_{rms}^2 = \frac{3PV}{M}$Comparing this with the equation of a straight line $y = mx + C$,where $y = u_{rms}^2$ and $x = PV$:$y = \left(\frac{3}{M}\right)x + 0$Thus,the slope $(m)$ is $\frac{3}{M}$.

At $T$ $(K)$,the $P$,$V$ and $u_{rms}$ of $1$ mole of an ideal gas were measured. The following graph is obtained. What is its slope $(m)$? ($x$-axis $= PV$; $y$-axis $= u_{rms}^2$,$M =$ Molar mass)

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