A gas bubble with a volume of 1.0 mL originates at the bottom of a lake where the pressure is 3.00 atm. Calculate its volume when the bubble reaches the surface of the lake where the pressure is 695 torr , assuming that the temp doesn't change.
I got 0.003 L, is that correct
Your answer is ok unless you want to be picky about the number of significant figures. You are allowed 2 (from the 2 in 1.0 mL). I get 3.28 mL which would round to 3.3 mL or 0.0033 L.
To solve this problem, we can use Boyle's Law, which states that the volume of a given amount of gas is inversely proportional to its pressure at a constant temperature.
Boyle's Law can be represented as:
P1 * V1 = P2 * V2
Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively.
Given:
- Initial volume (V1) = 1.0 mL = 0.001 L
- Initial pressure (P1) = 3.00 atm
- Final pressure (P2) = 695 torr
Now, we can plug these values into the Boyle's Law equation:
3.00 atm * 0.001 L = 695 torr * V2
Next, let's convert the torr pressure to atm:
1 atm = 760 torr
695 torr = 695/760 atm ≈ 0.9145 atm
0.001 L = 0.9145 atm * V2
Now, solve for V2:
V2 = 0.001 L / 0.9145 atm
V2 ≈ 0.001092 L
So, the volume of the bubble when it reaches the surface of the lake is approximately 0.001092 L. Therefore, your answer of 0.003 L is not correct.
To solve this problem, you can use Boyle's Law, which states that the product of the initial volume (V1) and initial pressure (P1) is equal to the product of the final volume (V2) and final pressure (P2), as long as the temperature remains constant.
The equation for Boyle's Law is:
P1 * V1 = P2 * V2
Given:
Initial volume (V1) = 1.0 mL
Initial pressure (P1) = 3.00 atm
Final pressure (P2) = 695 torr
First, let's convert the given values to a consistent unit. Since 1 atm = 760 torr, we can convert the final pressure to atm:
Final pressure (P2) = 695 torr * (1 atm / 760 torr) = 0.9145 atm
Now we can solve for the final volume (V2):
P1 * V1 = P2 * V2
(3.00 atm) * (1.0 mL) = (0.9145 atm) * V2
V2 = (3.00 atm * 1.0 mL) / (0.9145 atm)
V2 = 3.2732 mL
To convert mL to L:
V2 = 3.2732 mL * (1 L / 1000 mL) = 0.0032732 L
So, the final volume when the gas bubble reaches the surface of the lake is approximately 0.003 L. Therefore, your answer of 0.003 L is correct.