# 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.

• It is not $PV=mRT$, but rather, $PV=nRT$, where $n$ is the amount of substance of gas (in moles). – Douglas Fir Oct 1 '15 at 1:11
• This is a quote copped word for word from my textbook, which is why the fact that it contradicts the actual formula confuses me. The textbook is called "Engineering Fundamentals: An Introduction to Engineering" 5th edition. It is problem 6.31. – FabulousGlobe Oct 1 '15 at 1:27

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.

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.

• Oh, engineers and their silly notation... this is probably the right answer, then. (+1) – Chappers Oct 1 '15 at 1:35
• @Louis I skimmed through the book and found a section that mentions this formula: $R =$ gast constant $\frac{J}{kg*K} or \frac{ft * lb}{lbm *R}$ Thank you for everything, you were right. I have learned a valuable lesson along the lines of, "Always check the other sections of the textbook, regardless of what the index says." – FabulousGlobe Oct 1 '15 at 1:45

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.

• I wasn't the one that wrote that formula, it is a quote directly from my textbook, which is why I am so confused. In case you are curious, this is "Engineering Fundamentals: An Introduction to Engineering" 5th edition. Problem 6.31. – FabulousGlobe Oct 1 '15 at 1:29