By using the ideal gas equation $PV = \mu RT$,write the value,unit,and dimension of $R$.

(N/A) The ideal gas equation is given by $PV = \mu RT$.
Rearranging for $R$,we get $R = \frac{PV}{\mu T}$.
At standard temperature and pressure $(STP)$,for $1 \text{ mole}$ of an ideal gas $(\mu = 1)$:
$P = 1.013 \times 10^{5} \text{ N/m}^2$
$V = 22.4 \times 10^{-3} \text{ m}^3$
$T = 273.15 \text{ K}$
Substituting these values:
$R = \frac{1.013 \times 10^{5} \times 22.4 \times 10^{-3}}{1 \times 273.15} \approx 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$.
Unit of $R$: $\text{J mol}^{-1} \text{ K}^{-1}$ or $\text{N m mol}^{-1} \text{ K}^{-1}$.
Dimension of $R$:
Since $R = \frac{PV}{\mu T}$,the dimension is $\frac{[M^1 L^{-1} T^{-2}] [L^3]}{[mol] [K]} = [M^1 L^2 T^{-2} K^{-1} mol^{-1}]$.