A balloon is filled with 0.75 L of helium gas at 1 atm pressure and 25 degrees Celsius. The balloon is released and floats upward in the atmosphere. The temperature drops to 23 degrees Celsius and the balloon expands to 1.50 L. What is the pressure at this altitude? Assume the gas behaves ideally.
To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the absolute temperature.
First, let's convert the temperatures to Kelvin. We add 273 to convert Celsius to Kelvin.
Initial temperature (T1) = 25°C + 273 = 298 K
Final temperature (T2) = 23°C + 273 = 296 K
Now, let's solve for the initial number of moles (n1) using the initial conditions.
Initial volume (V1) = 0.75 L
Pressure (P1) = 1 atm
R = 0.0821 L·atm/(mol·K)
Using the ideal gas law equation, we can rearrange it to solve for the number of moles:
n1 = (P1 * V1) / (R * T1)
n1 = (1 atm * 0.75 L) / (0.0821 L·atm/(mol·K) * 298 K)
Calculating the above expression gives us:
n1 ≈ 0.0308 mol (rounded to four decimal places)
Now, let's solve for the final pressure (P2) using the final conditions.
Final volume (V2) = 1.50 L
R = 0.0821 L·atm/(mol·K)
Number of moles (n2) remains the same as n1 from the initial conditions.
Using the ideal gas law equation, we can rearrange it to solve for the pressure:
P2 = (n2 * R * T2) / V2
P2 = (0.0308 mol) * (0.0821 L·atm/(mol·K)) * (296 K) / (1.50 L)
Calculating the above expression gives us:
P2 ≈ 0.502 atm (rounded to three decimal places)
Therefore, at this altitude, the pressure is approximately 0.502 atm.
To find the pressure at this altitude, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = moles of the gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
First, let's convert the temperatures from Celsius to Kelvin:
25 degrees Celsius + 273.15 = 298.15 K
23 degrees Celsius + 273.15 = 296.15 K
Since the amount of gas (moles) in the balloon remains constant, we can rewrite the ideal gas law equation as:
(P1)(V1) / (T1) = (P2)(V2) / (T2)
P1 = 1 atm (initial pressure)
V1 = 0.75 L (initial volume)
T1 = 298.15 K (initial temperature)
V2 = 1.50 L (final volume)
T2 = 296.15 K (final temperature)
Now, plug in the values and solve for P2 (the final pressure):
(1 atm)(0.75 L) / (298.15 K) = (P2)(1.50 L) / (296.15 K)
P2 = [(1 atm)(0.75 L)(296.15 K)] / [(1.50 L)(298.15 K)]
P2 ≈ 0.996 atm
Therefore, the pressure at this altitude is approximately 0.996 atm.
(P1V1)/T1 = (P2V2)/T2
Don't forget to use T in Kelvin.