Suppose a balloon is filled so that its volume is 2.00 L when the pressure is 1.10 atm and the temperature is 300k. What volume will it occupy if it rises to an elevation where the pressure is 418mm Hg and the temperature is 200k?
See your previous post. Make sure P1 and P2 are in the same units. 1 atm = 760 mm. Post your work if you get stuck.
Cry
To solve this problem, we can use the combined gas law formula, which relates the initial and final conditions of pressure, volume, and temperature. The formula is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Let's substitute the given values into the formula:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
(1.10 atm * 2.00 L) / (300 K) = (418 mmHg * V2) / (200 K)
Now let's solve for V2:
V2 = (1.10 atm * 2.00 L * 200 K) / (418 mmHg * 300 K)
V2 ≈ 1.89 L
Therefore, the volume that the balloon will occupy when it rises to the new elevation is approximately 1.89 L.
To solve this problem, we can use the ideal gas law equation, which is given by:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
First, we need to convert the pressure and temperature to the correct units. The pressure is given in mm Hg, so we need to convert it to atm:
1 atm = 760 mm Hg
So, the pressure at the new elevation is:
418 mm Hg / 760 mm Hg/atm = 0.549 atm
The temperature is given in Kelvin, which is the correct unit for the ideal gas law.
Next, we can set up a ratio using the ideal gas law equation:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
Where the subscripts ₁ and ₂ represent the initial and final conditions, respectively.
Plugging in the given values:
(1.10 atm * 2.00 L) / (300 K) = (0.549 atm * V₂) / (200 K)
Now, we can solve for V₂:
(2.20 atm * L) / (300 K) = (0.549 atm * V₂) / (200 K)
(2.20 atm * L * 200 K) = (300 K * 0.549 atm * V₂)
440 atm-L-K = 164.7 atm-L-K * V₂
V₂ = (440 atm-L-K) / (164.7 atm-L-K)
V₂ ≈ 2.67 L
Therefore, the balloon will occupy approximately 2.67 liters at the new elevation.