Among terrestrial animals, the African elephant has one of the largest lung capacities, estimated at 300 L (for comparison, human lung capacity is around 5.0 L). An African elephant is passing through the highland region of Kenya, where the pressure is 092 atm and the temperature is 32^circ C(90^circ F) Show equations and constants v Calculate the mass of air that can be held in the elephant's lungs under the conditions given (the molar mass of air is 29g/mol) Choose 1 answer:

To calculate the mass of air that can be held in the elephant's lungs, we can use the ideal gas law equation:<br /><br />\[ PV = nRT \]<br /><br />Where:<br />- \( P \) is the pressure in atmospheres (atm)<br />- \( V \) is the volume in liters (L)<br />- \( n \) is the number of moles of gas<br />- \( R \) is the ideal gas constant, which is \( 0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \)<br />- \( T \) is the temperature in Kelvin (K)<br /><br />First, convert the temperature from Celsius to Kelvin:<br /><br />\[ T = 32^\circ C + 273.15 = 305.15 \, \text{K} \]<br /><br />Given:<br />- Pressure \( P = 0.92 \, \text{atm} \)<br />- Volume \( V = 300 \, \text{L} \)<br /><br />Now, solve for \( n \) (the number of moles of air):<br /><br />\[ n = \frac{PV}{RT} = \frac{(0.92 \, \text{atm})(300 \, \text{L})}{(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K})(305.15 \, \text{K})} \]<br /><br />Calculate \( n \):<br /><br />\[ n \approx \frac{276}{25.055115} \approx 11.01 \, \text{moles} \]<br /><br />Finally, calculate the mass of air using the molar mass of air (\( 29 \, \text{g/mol} \)):<br /><br />\[ \text{Mass} = n \times \text{molar mass} = 11.01 \, \text{moles} \times 29 \, \text{g/mol} \]<br /><br />\[ \text{Mass} \approx 319.29 \, \text{g} \]<br /><br />Therefore, the mass of air that can be held in the elephant's lungs under the given conditions is approximately 319.29 grams.