A chamber has a volume of 17 m3 and operates at a pressure of 4.66 atm. What volume in cubic meters would the air in the chamber occupy if the pressure were changed to 1.64 atm at constant temperature?
48.3m^3
P1V1 = P2V2
To solve this problem, we can use the combined gas law, which states:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature (assuming constant temperature in this case)
P2 = final pressure
V2 = final volume
T2 = final temperature (assuming constant temperature in this case)
We are given:
P1 = 4.66 atm
V1 = 17 m³
P2 = 1.64 atm
T1 = T2 (constant temperature)
Let's solve for V2.
First, rewrite the equation as:
(V1 * P1) / T1 = (V2 * P2) / T2
Since the temperature is constant (T1 = T2), we can simplify the equation to:
(V1 * P1) = (V2 * P2)
Now plug in the values we have:
(17 m³ * 4.66 atm) = (V2 * 1.64 atm)
Now, isolate V2 by dividing both sides of the equation by 1.64 atm:
(17 m³ * 4.66 atm) / 1.64 atm = V2
V2 = (17 m³ * 4.66 atm) / 1.64 atm
Simplifying the equation gives:
V2 ≈ 48.09 m³
Therefore, the air in the chamber would occupy approximately 48.09 cubic meters if the pressure were changed to 1.64 atm at constant temperature.