To solve this problem, we can use the ideal gas law, which states:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Given: Initial conditions: Temperature (T1) = 0.0 °C = 273.15 K Pressure (P1) = 1.00 atm Volume (V1) = 2.24 L Number of moles (n1) = unknown
Final conditions: Temperature (T2) = 109 °C = 382.15 K Pressure (P2) = 1.00 atm (held constant) Number of moles (n2) = 0.10 mol
First, let's calculate the initial number of moles of hydrogen gas (H2) in the balloon using the ideal gas law:
P1V1 = n1RT1
n1 = (P1V1) / (RT1)
Substituting the values:
n1 = (1.00 atm * 2.24 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
n1 = 0.1079 mol
Now, we can calculate the final volume of the gas when 0.10 mol of helium (He) is added and the temperature is raised to 109 °C while the pressure and amount of gas are held constant. Since the pressure and number of moles are constant, we can use the combined gas law:
(P1V1) / (n1T1) = (P2V2) / (n2T2)
Substituting the known values:
(1.00 atm * 2.24 L) / (0.1079 mol * 273.15 K) = (1.00 atm * V2) / (0.10 mol * 382.15 K)
Simplifying the equation and solving for V2 (final volume):
V2 = [(1.00 atm * 2.24 L) / (0.1079 mol * 273.15 K)] * (0.10 mol * 382.15 K)
V2 ≈ 1.737 L
Therefore, the final gas volume, when 0.10 mol of helium is added and the temperature is raised to 109 °C while the pressure and amount of gas are held constant, is approximately 1.737 L.