To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles (which remains constant in this case)
R = ideal gas constant
T = temperature
Since we are assuming that the number of moles remains constant and the bag is sealed and flexible, we can simplify the equation to:
P1V1/T1 = P2V2/T2
Where:
P1 = initial pressure (765 mmHg)
V1 = initial volume (4.85 L)
T1 = initial temperature (25.0 °C)
P2 = final pressure (442 mmHg)
V2 = final volume (what we want to find)
T2 = final temperature (5.00 °C)
Now we can plug in the values and solve for V2:
(765 mmHg)(4.85 L)/(25.0 + 273.15) K = (442 mmHg)V2/(5.00 + 273.15) K
Here, we have converted the temperatures to Kelvin by adding 273.15 to the Celsius temperatures.
Solving for V2:
(765 mmHg)(4.85 L)/(298.15 K) = (442 mmHg)V2/(278.15 K)
V2 = (765 mmHg)(4.85 L)/(298.15 K) * (278.15 K)/(442 mmHg)
V2 = 3.97 L
Therefore, the volume of air in the plastic bag at the top of the mountain would be approximately 3.97 L.
Question 8
A sealed flexible plastic bag contains a volume of 4.85 L of air at 25.0 °C and a pressure of 765 mmHg at sea level.
Assuming that the bag does not break or leak, what will be its volume if it is transported to the top of a mountain where the pressure is 442 mmHg and the temperature is 5.00 °C?
1 answer