## 16.4 Gravitational Effects on Ideal Gas Model This section discusses the role of gravity in the ideal gas model, differentiating its significance based on scale. While gravity is negligible at laboratory scales, it becomes crucial for the stability of large celestial bodies like stars. It also introduces the concept of hydrostatic equilibrium and provides an example demonstrating the use of the Van der Waals equation. ### Gravitational Effects on Ideal Gas Model At the **laboratory scale**, gravity's influence on ideal gases is considered much weaker than other factors such as the kinetic energy of gas molecules and gas pressure. Therefore, it is typically ignored in ideal gas models for such scales. At a **large scale**, such as within a star or a very tall planetary atmosphere, gravity plays a significant and important role in their stability and density distribution. Accounting for gravity at this scale requires more complex models beyond the simple ideal gas model. For a deeper understanding of how gravity acts as a field, see . ### Hydrostatic Equilibrium This is a state of balance observed in stars. It occurs when the net outward pressure generated by factors such as heat from fusion, radiation pressure, and degeneracy pressure (due to quantum effects) is precisely counterbalanced by the inward force of gravity. This equilibrium is essential for maintaining the stability of stars. The strength of this inward pull depends on the mass and radius of the celestial body. ### Van der Waals Equation While the ideal gas model assumes point particles with no intermolecular forces (see ), real gases behave differently. The Van der Waals equation is a modified equation of state for real gases, accounting for the finite size of gas molecules and the attractive forces between them. The formula is: $$\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T$$ Where: - $P$ = pressure - $V$ = volume - $n$ = number of moles - $R$ = ideal gas constant - $T$ = temperature - $a$ = constant accounting for intermolecular attractive forces - $b$ = constant accounting for the volume occupied by gas molecules ## Worked Example **Question:** Calculate the pressure exerted by one mole of a gas at room temperature 300 K with fixed volume 0.022 m³, using the Van der Waals equation. Given: $a = 3.59\ \text{L}^2\cdot\text{atm}\cdot\text{mol}^{-2}$, $b = 0.0427\ \text{L}\cdot\text{mol}^{-1}$, and $R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$. **Solution:** **Given:** $T = 300\ \text{K}$, $n = 1\ \text{mol}$, $V = 0.022\ \text{m}^3 = 22\ \text{L}$, $a = 3.59\ \text{L}^2\cdot\text{atm}\cdot\text{mol}^{-2}$, $b = 0.0427\ \text{L}\cdot\text{mol}^{-1}$, $R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ **Van der Waals equation for pressure:** $$P=\frac{n R T}{V-n b}-\frac{a n^{2}}{V^{2}}$$ **Substituting values:** $$P=\frac{1 \times 0.0821 \times 300}{22-(1)(0.0427)}-\frac{3.59 \times (1)^{2}}{(22)^{2}}$$ **Result:** $P \approx 1.1146\ \text{atm}$