Question

13\. A balloon filled with helium has a volume of \(1.28 \times 10^{3} \mathrm{~L}\) at sea level where the pressure is 0.998 atm and the temperature is \(31^{\circ} \mathrm{C}\). The balloon is taken to the top of a mountain where the pressure is 0.753 atm and the temperature is \(-25^{\circ} \mathrm{C}\). What is the volume of the balloon at the top of the mountain?

Short answer

Answer: The volume of the helium balloon at the top of the mountain is approximately \(1.51 * 10^{3} \text{ L}\).

Step-by-step solution

Convert temperatures to Kelvin

To convert the temperatures from Celsius to Kelvin, add 273.15 to the given Celsius values. Initial temperature at sea level: \(31^{\circ}C = 31 + 273.15 = 304.15 K\) Final temperature at mountain top: \(-25^{\circ}C = -25 + 273.15 = 248.15 K\)

Set up the Ideal Gas Law equation for both conditions

Using the Ideal Gas Law (PV = nRT), we will have two equations for the two different conditions: At sea level: \(P_{1}V_{1} = nR T_{1}\) At mountain top: \(P_{2}V_{2} = nR T_{2}\)

Solve for the final volume

Since the number of moles of gas (n) and the Ideal Gas constant (R) do not change, we can equate the right sides of both equations: \(nR T_{1} = nR T_{2}\) Now, substitute the given values and the temperatures in Kelvin into the equation and solve for the final volume: \(P_{1}V_{1} / T_{1} = P_{2}V_{2} / T_{2}\) \(V_{2} = (P_{1}V_{1})(\frac{T_{2}}{T_{1}})/(P_{2})\) \(V_{2} = (0.998 \text{ atm} * 1.28*10^{3} \text{ L})(\frac{248.15 \text{ K}}{304.15 \text{ K}})/(0.753 \text{ atm})\) \(V_{2} = \frac{(0.998 * 1.28*10^{3})(248.15)}{(304.15)(0.753)}\) \(V_{2} ≈ 1.51 * 10^{3} \text{ L}\) So the volume of the balloon at the top of the mountain is approximately \(1.51 * 10^{3} \text{ L}\).