Problem Statement
Derive the barometric formula for an isothermal ideal gas atmosphere.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Consider a horizontal slab of gas at height $h$, thickness $dh$. Pressure balance:
Step 2 — Apply the relevant physical law or equation: $$dp = -\rho g\,dh$$
Step 3 — Solve algebraically for the unknown: Using $\rho = pM/(RT)$ from the ideal gas law:
Step 4 — Substitute numerical values with units: $$\frac{dp}{p} = -\frac{Mg}{RT}dh$$
Step 5 — Compute and check the result: Integrating with $p(0)=p_0$:
Step 6: $$\boxed{p(h) = p_0\exp\!\left(-\frac{Mgh}{RT}\right)}$$
Worked Calculation
$$dp = -\rho g\,dh$$
$$\frac{dp}{p} = -\frac{Mg}{RT}dh$$
$$\boxed{p(h) = p_0\exp\!\left(-\frac{Mgh}{RT}\right)}$$
Consider a horizontal slab of gas at height $h$, thickness $dh$. Pressure balance:
$$dp = -\rho g\,dh$$
Using $\rho = pM/(RT)$ from the ideal gas law:
$$\frac{dp}{p} = -\frac{Mg}{RT}dh$$
Integrating with $p(0)=p_0$:
$$\boxed{p(h) = p_0\exp\!\left(-\frac{Mgh}{RT}\right)}$$
The characteristic scale height is $H = RT/(Mg)$. For air at 273 K: $H \approx 8.0\ \text{km}$.
Answer
$$\boxed{p(h) = p_0\exp\!\left(-\frac{Mgh}{RT}\right)}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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