Assuming that atmospheric temperature decreases with increasing altitude at a uniform rate of \(0.0065 \mathrm{~K} \cdot \mathrm{m}^{-1}\), determine the atmospheric pressure at an altitude of \(7.5 \mathrm{~km}\) if the temperature and pressure at sea level are \(15^{\circ} \mathrm{C}\) and \(101.5 \mathrm{kPa}\) respectively. \(\left(R=287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\right)\)

Convert the given sea level temperature from Celsius to Kelvin. The formula for this conversion is \(T(\text{K}) = T(^{\text{∘}C}) + 273.15\)Given:\[T_0 = 15^{\text{∘}C} = 15 + 273.15 = 288.15 \text{K}\]

Use the temperature lapse rate to calculate the temperature at the given altitude. The formula is \( T = T_0 - L h \) where \(L\) is the lapse rate (\(0.0065 \text{ K }\text{m}^{-1}\)), \(h\) is the altitude\[ T = 288.15 \text{K} - (0.0065 \text{K}/\text{m}) \times 7500 \text{m} \]\[ T = 288.15 \text{K} - 48.75 \text{K} \]\[ T = 239.4 \text{K} \]

Plug the values into the barometric formula to find the pressure at altitude. The formula is \[ P = P_0 \times \bigg(1 - \frac{Lh}{T_0} \bigg)^{\frac{gM}{RL}} \]Given:\(P_0 = 101.5 \text{kPa}\), \(L = 0.0065 \text{ K }\text{m}^{-1}\), \(h = 7500 \text{m}\), \(T_0 = 288.15 \text{K}\), \(R = 287 \text{ J }\text{kg}^{-1}\text{K}^{-1} \), and gravitational constant \( g = 9.81 \text{m}/\text{s}^2 \), and molar mass of dry air \( M = 0.029 \text{kg}\text{mol}^{-1} \).\[ P = 101.5 \text{kPa} \times \bigg( 1 - \frac{0.0065 \times 7500}{288.15} \bigg)^{\frac{9.81 \times 0.029}{287 \times 0.0065}} \]\[ P = 101.5 \text{kPa} \times \bigg(1 - 0.169119 \bigg)^{5.255877} \]\[ P = 101.5 \text{kPa} \times (0.830881)^{5.255877} \]\[ P \approx 101.5 \text{kPa} \times 0.498 \]\[ P \approx 50.54 \text{kPa} \]