# Natural log of a pressure is ?

• ChrisW
In summary: P1 by using the definition of pressure and solving for T1 using the vaporization temperature and the ideal gas law.
ChrisW
Ok so i was doing a problem involving finding the pressure of mercury at its boiling point (630.05K) using Troutons rule and the final answer seems a bit strange to me.

Integration of the Clausius-Clapeyron Equation:

$${ln(\frac{{P}_{2}}{{P}_{1}}) = -\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2-\frac{1}{T}_1)$$

So... you end up with a number, but it's unitless? What am I missing? Thanks.

Troutons rule states that at standard T and P:

$${\Delta}_{vap}\overline{S} \approx {88} {J}\cdot{K}^{-1}\cdot{mol}^{-1}$$

and the change in entropy is related to the change in enthalpy by:

$${\Delta}_{vap}\overline{H}={\Delta}_{vap}\overline{S}\cdot{T}_{vap}$$

ChrisW said:
(snip)Integration of the Clausius-Clapeyron Equation:

$${ln(\frac{{P}_{2}}{{P}_{1}}) = -\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2-\frac{1}{T}_1)$$

So... you end up with a number, but it's unitless?

Correct
What am I missing?

Nothing --- well, maybe you should take another peek at the expression (P2/P1).

another hint is that you can never take logs or exponentials with anything that has units. it simply makes no sense.

gravenewworld said:
another hint is that you can never take logs or exponentials with anything that has units. it simply makes no sense.
Yes, this was the heart of my question. We use Troutons rule to determine the pressure at the vaporization temperature, yet to do this we must take the exponential of the equation. So, for instance:

$${ln(\frac{{P}_{2}}{{P}_{1}}) = -\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2 -\frac{1}{T}_1)$$

$${\frac{{P}_{2}}{{P}_{1}} = e^{-\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2 -\frac{1}{T}_1)}$$

$${P}_{2} = e^{-\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2 -\frac{1}{T}_1)}{P}_{1}$$

Are units retained after taking the exponential of $ln(\frac{{P}_2}{{P}_1})$? I mean I understand you're still taking the exponential of a unitless number because it's pascals/pascals, but my instructure mentioned that there was a special circumstance that kept the units around... Perhaps it is just simple math and he's just confusing us all :rofl:.

I also realize that the title of my post is misleading, sorry about that. ;)

Are units retained after taking the exponential of $ln(\frac{{P}_2}{{P}_1})$
? I mean I understand you're still taking the exponential of a unitless number because it's pascals/pascals, but my instructure mentioned that there was a special circumstance that kept the units around... Perhaps it is just simple math and he's just confusing us all .

if you take the exponential you will still have a unitless equation. like you said you will have p2/p1 so units will cancel out. Sure you can mutiply both sides by p1 to get p2=e^(stuff)p1 and you have an equation with pascals on the left and pascals on the right. But that just means you can still cancel out units. Don't get confused by playing around with the algebra. No matter how you play around with the equation, the units will always end up canceling out. Just ask yourself what is the ln(cm) or ln(bars) or e^cm or e^J. These unit quantities make no sense. I am 99.9% sure there are absolutely no cases where you can take the log or exponential of a quantity that has units.

I was not confused by the algebra at all. Really what I wanted to find out was that my answer, which I obtained from the very equation for P2 I posted above, was correctly stated as a pressure. The basic algebra of the problem made sense to me, however my instructor circled the units on my answer and noted in class that the reason was due to the logarithm. I wanted to see if I could figure it out on my own and I came to the conclusion that my answer was right, just wanted to make sure I wasn't making a mistake.

Thanks.

I really don't see the problem. You get an expression P2/P1 that is unitless for obvious reasons - the units cancel out on division. If P2/P1 = x then P2 = x times P1, P1 having a (arbitrary) unit of pressure like bar, kPa, atm, psi etc. and x being unitless. The only thing I wonder about is how to get P1 and T1... :uhh:

i didn't see the problem either! i took my paper back and received full credit .

as for finding P1 and T1, it's just more algebra.

## What is the natural log of a pressure?

The natural log of a pressure is a mathematical function that represents the logarithm of a pressure value in the base of the mathematical constant e (approximately equal to 2.71828). It is denoted as ln(P) or loge(P), where P is the pressure value.

## What is the significance of taking the natural log of a pressure?

The natural log of a pressure is commonly used in scientific and mathematical calculations to simplify equations and make them easier to solve. It can also help to convert a non-linear relationship between pressure and another variable into a linear relationship, making it easier to analyze and interpret.

## How is the natural log of a pressure different from the logarithm of a pressure in other bases?

The natural log of a pressure, denoted as ln(P), is specifically calculated using the base e, while other logarithmic functions use different bases such as 10 or 2. The natural log is often preferred in scientific calculations because of its unique mathematical properties and its relationship to the natural world.

## What is the range of values for the natural log of a pressure?

The natural log of a pressure can have a range of values from negative infinity to positive infinity. However, in practical applications, it is typically used for positive pressure values only, as negative pressure values do not have physical meaning.

## How is the natural log of a pressure related to other scientific concepts?

The natural log of a pressure is closely related to other scientific concepts such as the ideal gas law, thermodynamics, and fluid mechanics. It is often used in these fields to model and analyze the behavior of gases and fluids under different conditions.

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