A sample of air occupies 3.8 L when the pressure is 1.2 atm.
(a) What volume does it occupy at 7.3 atm?
(b) What pressure is required in order to compress it to 0.050 L? (The temperature is kept constant.)
I already know the answers to each question, but do not understand how to solve for it
A sample of air occupies 3.8L when the pressure is 1.2 atm, assuming the temperature remains constant what volume does it occupy at 6.6 atm
To solve both questions, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure, as long as the temperature remains constant.
Boyle's Law formula: P1 * V1 = P2 * V2
where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
(a) To find the volume at 7.3 atm:
Step 1: Write down the given values:
P1 = 1.2 atm
V1 = 3.8 L
P2 = 7.3 atm
Step 2: Substitute the values into the formula and solve for V2:
P1 * V1 = P2 * V2
(1.2 atm) * (3.8 L) = (7.3 atm) * V2
(1.2 L * atm) / (7.3 atm) = V2
V2 ≈ 0.197 L
Therefore, the volume is approximately 0.197 L at a pressure of 7.3 atm.
(b) To find the pressure required to compress it to 0.050 L:
Step 1: Write down the given values:
P1 = 1.2 atm
V1 = 3.8 L
V2 = 0.050 L
Step 2: Substitute the values into the formula and solve for P2:
P1 * V1 = P2 * V2
(1.2 atm) * (3.8 L) = P2 * (0.050 L)
(1.2 L * atm) / (0.050 L) = P2
P2 ≈ 28.8 atm
Therefore, a pressure of approximately 28.8 atm is required to compress the air to 0.050 L.
To solve both parts of this question, you can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature is kept constant.
Boyle's Law can be expressed as: P1 × V1 = P2 × V2
Let's solve part (a) first:
(a) What volume does the air occupy at 7.3 atm?
We are given:
P1 = 1.2 atm (initial pressure)
V1 = 3.8 L (initial volume)
P2 = 7.3 atm (new pressure)
V2 = unknown (new volume)
Using Boyle's Law, we can write:
P1 × V1 = P2 × V2
Substituting the given values:
(1.2 atm) × (3.8 L) = (7.3 atm) × (V2)
Now, rearrange the equation to solve for V2:
V2 = (P1 × V1) / P2
Plug in the values:
V2 = (1.2 atm × 3.8 L) / 7.3 atm
V2 = 0.624 L (rounded to three decimal places)
Therefore, the volume of air at 7.3 atm is approximately 0.624 L.
Now, let's move on to part (b):
(b) What pressure is required to compress the air to 0.050 L?
We are given:
P1 = 1.2 atm (initial pressure)
V1 = 3.8 L (initial volume)
P2 = unknown (new pressure)
V2 = 0.050 L (compressed volume)
Apply Boyle's Law:
P1 × V1 = P2 × V2
Substituting the given values:
(1.2 atm) × (3.8 L) = (P2) × (0.050 L)
Isolate P2:
P2 = (P1 × V1) / V2
Plug in the values:
P2 = (1.2 atm × 3.8 L) / 0.050 L
P2 = 91.2 atm (rounded to one decimal place)
Therefore, the pressure required to compress the air to 0.050 L is approximately 91.2 atm.