A hot air balloon with a volume of 3.20 x 103 L and 120 moles of air is at a temperature of 25 °C. The air is heated to 65°C and 250 moles of air is added. What will the new volume of the balloon be?
To solve this problem, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
Let's break down the given information:
Initial volume (V1): 3.20 x 10^3 L
Initial moles of air (n1): 120 mol
Initial temperature (T1): 25 °C
Now let's calculate the final volume (V2):
1. Convert the temperatures to Kelvin:
Initial temperature (T1) = 25 °C + 273.15 = 298.15 K
Final temperature (T2) = 65 °C + 273.15 = 338.15 K
2. Calculate the total number of moles of air:
Final moles of air (n2) = n1 + 250 mol = 120 mol + 250 mol = 370 mol
3. Use the ideal gas law equation to find the final volume:
P1V1 = n1RT1
P2V2 = n2RT2
Since the pressure (P) and the ideal gas constant (R) are constant, we can set the two equations equal to each other:
P1V1 = P2V2
Now we can rearrange the equation to solve for the final volume (V2):
V2 = (P1V1)/(P2) = (n1RT1)/(n2T2)
Substituting the given values:
V2 = (120 mol * 0.0821 atm L/mol K * 298.15 K) / (370 mol * 338.15 K)
Calculating the expression:
V2 = 0.01972 L
Therefore, the new volume of the balloon will be approximately 0.01972 L.