A hot air balloon is filled with 1.51 × 106 L of an ideal gas on a cool morning (11 °C). The air is heated to 103 °C. What is the volume of the air in the balloon after it is heated? Assume that none of the gas escapes from the balloon.
the volume is directly proportional to the ABSOLUTE (ºKelvin) temperature
... K = C + 273
v = 1.51E6 L * [(103 + 273) / (11 + 273)]
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant
T = temperature of the gas
Since the amount of gas inside the balloon remains constant, we can consider n and R as constant. Therefore, we can rewrite the equation as:
V1/T1 = V2/T2
Where:
V1 = initial volume of the gas (1.51 × 10^6 L)
T1 = initial temperature of the gas (11 °C + 273.15 K)
V2 = final volume of the gas (what we want to find)
T2 = final temperature of the gas (103 °C + 273.15 K)
Let's substitute the values into the equation and solve for V2:
(V1/T1) = (V2/T2)
(1.51 × 10^6 L) / (11 °C + 273.15 K) = V2 / (103 °C + 273.15 K)
Now, let's calculate:
(1.51 × 10^6 L) / (284.15 K) = V2 / (376.15 K)
V2 = (1.51 × 10^6 L) * (376.15 K) / (284.15 K)
V2 = 1,993,989.17 L
Therefore, the volume of the air in the balloon after it is heated is approximately 1,993,989.17 L.
To solve this problem, we need to use Charles's Law, which states that the volume of an ideal gas is directly proportional to its temperature, given that pressure and amount of gas remain constant. Charles's Law can be expressed as V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.
Given:
V1 = 1.51 × 10^6 L (initial volume)
T1 = 11 °C = 11 + 273 = 284 K (initial temperature)
T2 = 103 °C = 103 + 273 = 376 K (final temperature)
We can rearrange the equation as V2 = V1 * (T2 / T1) to find the final volume.
Substituting the values into the equation:
V2 = 1.51 × 10^6 L * (376 K / 284 K)
V2 = 1.51 × 10^6 L * 1.3246
Calculating the expression:
V2 = 2.010666 × 10^6 L
Therefore, the volume of the air in the balloon after it is heated is approximately 2.010666 × 10^6 L.