A spherical, helium-filled balloon of diameter \(800 \mathrm{~mm}\) is to be used to carry meteorological instruments to a height of \(6000 \mathrm{~m}\) above sea level. The instruments have a mass of \(60 \mathrm{~g}\) and negligible volume, and the balloon itself has a mass of \(100 \mathrm{~g}\). Assuming that the balloon does not expand and that atmospheric temperature decreases with increasing altitude at a uniform rate of \(0.0065 \mathrm{~K} \cdot \mathrm{m}^{-1}\), determine the mass of helium required. Atmospheric pressure and temperature at sea level are \(15^{\circ} \mathrm{C}\) and \(101 \mathrm{kPa}\) respectively; for air, \(R=287 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

The Ideal Gas Law is a fundamental equation for understanding the behavior of gases. The law is represented as \(PV = nRT\), where \(P\) represents pressure, \(V\) is volume, \(n\) is the number of moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. In many practical exercises, we rearrange the formula to solve for different variables. For example, to find the mass of a gas, we can use the form \(m = \frac{PV}{RT}\), where \(m\) is the mass and \(R\) is replaced by the specific gas constant for the particular gas (e.g., helium has \(R = 2077 \text{ J/(kg*K)} \). Understanding this law allows us to calculate how gases behave under different pressures, volumes, and temperatures.

Atmospheric pressure is the force exerted by the weight of air on the Earth's surface and objects at different altitudes. It decreases with altitude because there is less air above to exert force. At sea level, the standard atmospheric pressure is approximately 101 kPa. To calculate atmospheric pressure at a higher altitude, such as 6000 meters, we can use the barometric formula: \[ P = P_0 \bigg(1 - \frac{Lh}{T_0} \bigg)^{\frac{gM}{RL}} \] where \(P_0\) is the initial pressure, \(L\) is the temperature lapse rate, \(h\) is the altitude, \(T_0\) is the sea level temperature, \(g\) is the acceleration due to gravity, \(M\) is the molar mass of Earth's air, and \(R\) is the gas constant. Substituting these values, we find the pressure decreases significantly at higher altitudes.

Understanding density and volume is crucial for problems involving buoyancy and gas behavior. The volume of a sphere, such as a balloon, is calculated using \[ V = \frac{4}{3} \pi r^3 \] where \(r\) is the radius. For our 800 mm diameter balloon, the radius \(r\) is 0.4 meters, giving a volume of approximately 0.268 cubic meters. Density is the mass per unit volume, usually expressed as \( \rho = \frac{m}{V}\). Knowing the mass and volume of the gas inside the balloon helps determine its buoyancy, directly affecting whether it can carry additional instruments and rise to the desired altitude.

Temperature decreases with altitude at a rate known as the lapse rate, typically around 0.0065 K/m in the troposphere. To find the temperature at a given altitude, we use this relationship. Starting with the sea level temperature (15°C or 288.15 K), we can determine the temperature at 6000 meters: \[ T = 288.15 - (0.0065 \times 6000) = 249.15 \text{ K} \] This cooling effect is crucial for accurate pressure calculations and understanding how gas volume and density change with altitude, which affects balloon buoyancy and flight mechanics.