Im a little bit confused at the ending part of In(2)

Certainly! Let's break down that part of the calculation to clarify things.

### Context

We started with the barometric formula and simplified it by recognizing that at a certain height \( h \), the pressure \( P(h) \) is half the pressure at sea level:

\[
\frac{P(h)}{P_0} = \frac{1}{2} = e^{-\frac{Mgh}{R \cdot T(h)}}
\]

Taking the natural logarithm (ln) on both sides:

\[
\ln\left(\frac{1}{2}\right) = -\frac{Mgh}{R \cdot T(h)}
\]

This simplifies to:

\[
-\ln(2) = -\frac{Mgh}{R \cdot T(h)}
\]

The negative signs cancel out, leaving:

\[
\ln(2) = \frac{Mgh}{R \cdot T(h)}
\]

### Solving for \( h \)
We want to find the height \( h \), so we rearrange the equation:

\[
h = \frac{\ln(2) \cdot R \cdot T(h)}{M \cdot g}
\]

Since the temperature \( T(h) \) changes with height, we approximated it using the average temperature between the ground and a height of 1 km. The equation we solved was:

\[
h = \frac{\ln(2) \cdot R \cdot T_{\text{avg}}}{M \cdot g}
\]

Where:
- \( \ln(2) \) is the natural logarithm of 2, approximately 0.693.
- \( R \) is the universal gas constant (8.314 J/(mol·K)).
- \( T_{\text{avg}} \) is the average temperature in Kelvin.
- \( M \) is the molar mass of air (0.029 kg/mol).
- \( g \) is the acceleration due to gravity (9.81 m/s²).

Finally, after substituting these values, we calculated \( h \) to be approximately 6495 meters.

Let me know if you'd like further clarification or have more questions!

Explore the solution to the ln(2) problem in the barometric formula. Learn how to solve equations involving natural logarithms and barometric pressure. This problem covers key mathematical concepts such as the natural logarithm, barometric formula, and equation solving. Suitable for advanced high school students and above.

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