Hello, does anyone know how to solve these problems?
1. A bubble of CO₂ with a volume of 0.245 cm³ is released at the bottom of the ocean where the pressure is 388.7 kPa. What will the volume of the bubble be (in cubic centimeters) when it reaches the surface where the pressure is 765 mmHg?
2.A diver with a lung volume of 6.25 L at a pressure of 785 torr dives into the water until her lung volume is 5.87 L. What is the pressure of the water (in atmospheres) pushing on her at this depth?
1. (P1V1 = (P2V2)
Substitute and solve for the one unknown you have which is V2.
Note: I assume that since you don't know the temperature at the bottom of the ocean and you don't list the temperature at top that you don't expect any corrections to be made for the vapor pressure of water.
2. (P1V1) = (P2V2)
Substitute and solve for P2.
Post your work if you get stuck.
Thanks for the help!
*I also forgot to note that is says to round to 3 significant figures.
So I did the work with the formula that you wrote, but it says my answer is wrong. I tried rounding the answers in different ways, but it still was wrong. Here is my work:
1. 0.245*388.7=765*x
I got 0.124485, which I rounded to 0.124
2. 6.25*785=x*5.87
I got 835.81771, which I rounded to 836.
Do you know why these answers are incorrect?
Sure! I can help you solve these problems.
1. To solve the first problem, we need to use Boyle's Law, which states that the product of the pressure and volume of a gas is constant at constant temperature. The equation for Boyle's Law is:
P₁V₁ = P₂V₂
where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.
In this case, the initial volume (V₁) is 0.245 cm³, and the initial pressure (P₁) is 388.7 kPa. The final pressure (P₂) is 765 mmHg. However, we need to convert mmHg to kPa because the initial pressure is given in kPa. 1 mmHg is equal to 0.133 kPa.
Converting the final pressure to kPa:
765 mmHg × (0.133 kPa / 1 mmHg) = 101.945 kPa
Now we can plug in the values into the Boyle's Law equation and solve for V₂:
(388.7 kPa) × (0.245 cm³) = (101.945 kPa) × V₂
V₂ = (388.7 kPa × 0.245 cm³) / 101.945 kPa
V₂ ≈ 0.934 cm³
Therefore, the volume of the bubble when it reaches the surface will be approximately 0.934 cm³.
2. To solve the second problem, we can use the combined gas law, which relates the initial and final volumes, pressures, and temperatures of a gas. The equation for the combined gas law is:
(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂
where P₁ and V₁ are the initial pressure and volume, T₁ is the initial temperature, P₂ and V₂ are the final pressure and volume, and T₂ is the final temperature.
In this case, the initial volume (V₁) is 6.25 L, the initial pressure (P₁) is 785 torr, and the final volume (V₂) is 5.87 L. However, the problem does not provide the temperatures.
Assuming the temperature remains constant, we can cancel out the temperature terms from the equation, and the equation becomes:
(P₁ × V₁) = (P₂ × V₂)
Now, we can plug in the values given and solve for P₂:
(785 torr × 6.25 L) = (P₂ × 5.87 L)
P₂ = (785 torr × 6.25 L) / 5.87 L
P₂ ≈ 834.42 torr
Since the problem asks for the pressure in atmospheres, we need to convert torr to atmospheres. 1 atmosphere is equal to 760 torr.
Converting the pressure to atmospheres:
834.42 torr / 760 torr/atm ≈ 1.097 atm
Therefore, the pressure of the water pushing on the diver at this depth is approximately 1.097 atm.