How would you interpret this unit conversion question? The following question is copied word for word from my textbook, which is what causes me to be so confused about the contradiction that it implies. 
The question:
For gases under certain conditions, there is a relationship between
the pressure of the gas, its volume, and its temperature as given by what is commonly called the ideal gas law. The ideal gas law is:
PV = mRT
where 
P = Absolute pressure of the gas(Pa)
V = volume of the gas $m^3$ 
m = mass (kg) 
R = gas constant 
T = absolute temperature (kelvin). 
My Solution:
Solving this question goes leads me to an illogical conclusion: 
$\frac{PV}{mT} = R$ 
$\frac{(\frac{Kg}{m*s^2}) * m^3}{kg * K} = R$
$\frac{m^2}{s^2 * kelvin} = R$
But I know, from googling and prior experience that:
$R = \frac{joul}{mol * kelvin}$
$R = \frac{kg * m^2}{s^2 * mol * k}$
Somehow, I am missing a kilogram. 
 A: There is a universal gas constant (also called the molar gas constant) that can be used for any ideal gas in the equation $P V = n R T$.  There is also an individual gas constant (also called the engineering gas constant or the specific gas constant) for each gas or mixture (oxygen, air, etc) that can be used in the equation $P = \rho R_e T$  or  $P V = m R_e T$.  Your textbook should tell you which constant it means.  The book may have a table that provides values of the engineering gas constant for various gasses.
A: Notice, in the given equation $$PV=mRT\implies R=\frac{PV}{mT}$$ 
$m\ (kg)$ is the mass of gas & $R$ is called specific gas constant which has unit $\frac{joule}{kg\cdot K}$ as you have calculated. 
Now, the gas equation in term of Universal gas constant $\bar{R}$ is given as $$PV=n\bar{R}T\implies \bar{R}=\frac{PV}{nT}$$ where, $n$ is number of moles of the gas. & universal gas constant $\bar{R}$ has unit $\frac{joule}{mole\cdot K}$
If $M\ \frac{kg}{mole}$ is the molar mass of a given (specific) gas then specific gas constant $R$ & universal gas constant $\bar{R}$ are co-related as follows 
$$\color{red}{R=\frac{\bar {R}}{M}}$$
Hence,  setting the units of $\bar R$ & $M$, we get unit of specific gas constant  $$\large =\frac{\frac{joule}{mole\cdot K}}{\frac{kg}{mole}}$$
$$\large =\frac{\frac{kg\cdot m^2}{s^2\cdot mole\cdot K}}{\frac{kg}{mole}}=\frac{m^2}{s^2\cdot K}$$
Hence, your answer $\frac{m^2}{s^2\cdot kelvin}$ is correct because $R$ is a specific gas constant not universal gas constant.  
A: Notice that you're also missing a division by moles. The ideal gas law in physics and chemistry is written as
$$ PV=nRT, $$
where $n$ is the number of moles of the substance. Then the calculation works. Apparently your book uses a different convention, which it should specify.
