# Dimension of an Exponential with Dimension

My thermodynamics professor gives a formula for ideal gas like this: $$s _ { 2 } - s _ { 1 } = c _ \mathrm{ v , a v g } \ln \frac { T _ { 2 } } { T _ { 1 } } + R \ln \frac { v _ { 2 } } { v _ { 1 } }.$$ This is true and makes sense in dimension analysis because both $$\frac { T _ { 2 } } { T _ { 1 } }$$ and $$\frac { v _ { 2 } } { v _ { 1 } }$$ is dimensionless. But next comes the weird part:

For an isentropic process $$(\Delta S=0)$$ \begin{align} c _ v \ln \frac { T _ { 2 } } { T _ { 1 } } &=- R \ln \frac { v _ { 2 } } { v _ { 1 } }\\ \exp(c _ v \ln \frac { T _ { 2 } } { T _ { 1 } })&=\exp(- R \ln \frac { v _ { 2 } } { v _ { 1 } })\\ \left(\frac { T _ { 2 } } { T _ { 1 } }\right)^{c _ v}&=\left(\frac { V _ { 1 } } { V _ { 2 } }\right)^{R}\\ T^{C_V}V^R&=\mathrm{Const} \end{align}

Now, my problem is: all 4 elements in the LHS have (different) dimensions. How does this formula make sense or my professor is just wrong?

There are two important points here:

• $$c_v$$ and $$R$$ have the same units (namely, J/K), otherwise the quantities could not be added together;

• Even if the units aren't necessarily physically sensible in a given expression, as long as the units are the same on both sides, the equation is still logically consistent.

To show you that there is no problem here, we can write the end expression in a form that will give you sensible units, with little to no extra effort:

$$\left(\frac{T_2}{T_1}\right)^{c_v}=\left(\frac{V_1}{V_2}\right)^R\implies\left(\frac{T_2}{T_1}\right)^{c_v/R}=\frac{V_2}{V_1}$$

Since $$c_v$$ and $$R$$ have the same units, $$c_v/R$$ is dimensionless, so we can say:

$$VT^{c_v/R}=\mathrm{Const}$$

or equivalently

$$V^{R/c_v}T=\mathrm{Const}$$

It should be easy to see that the whole calculation can be done in terms of this dimensionless ratio $$c_v/R$$.

A constant need not be dimensionless. For example the speed of light is a constant with dimensions of velocity.

Exponents DO need to be dimensionless though. But you can always manage that by multiplying both sides of an equation by some constant with appropriate dimensions (1/heat capacity in your case) before exponentiating.