Respuesta :
Answer: 624.98 mL
Explanation:
According to the Ideal Gas Law:
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex] (1)
Where:
[tex]P_{1}=2.4 atm[/tex] is the pressure of the bubble under the surface
[tex]V_{1}=250 mL[/tex] is the volume of the bubble under the surface
[tex]T_{1}=15\°C + 273.15=288.15 K[/tex] is the temperature of the bubble under the surface
[tex]P_{2}=1 atm[/tex] is the pressure of the bubble at the surface
[tex]V_{2}[/tex] is the volume of the bubble at the surface
[tex]T_{2}=27\°C + 273.15=300.15 K[/tex] is the temperature of the bubble at the surface
So, we have to find [tex]V_{2}[/tex]:
[tex]V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}[/tex] (2)
[tex]V_{2}=\frac{(2.4 atm)(250 mL)(300.15 K)}{(288.15 K)(1 atm)}[/tex] (3)
Finally:
[tex]V_{2}=624.98 mL[/tex] This is the volume of the bubble when it reaches the surface
The volume of the bubble when it reaches the surface is about 625 mL
Further explanation
The basic formula of pressure that needs to be recalled is:
Pressure = Force / Cross-sectional Area
or symbolized:
[tex]\large {\boxed {P = F \div A} }[/tex]
P = Pressure (Pa)
F = Force (N)
A = Cross-sectional Area (m²)
Let us now tackle the problem !
In this problem , we will use Ideal Gas Law as follows:
Given:
Initial Volume = V₁ = 250 mL
Initial Pressure = P₁ = 2.4 atm
Initial Temperature = T₁ = 15 + 273 = 288 K
Final Pressure = P₂ = 1.0 atm
Final Temperature = T₂ = 27 + 273 = 300 K
Unknown:
Final Volume = V₂ = ?
Solution:
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
[tex]\frac{2.4(250)}{288} = \frac{1.0V_2}{300}[/tex]
[tex]\frac{600}{288} = \frac{1.0V_2}{300}[/tex]
[tex]\frac{25}{12} = \frac{V_2}{300}[/tex]
[tex]V_2 = \frac{25}{12} \times 300[/tex]
[tex]V_2 = 625 \texttt{ mL}[/tex]
[tex]\texttt{ }[/tex]
Learn more
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Answer details
Grade: High School
Subject: Physics
Chapter: Pressure
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure
