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v
Convert the given temperatures from celsius to kelvin since we are dealing with gas.
To convert to kelvin, add 273.15 to the temperature in celsius.
T1 = 22 + 273.15 = 295.15 k
T2 = 4 + 273.15 = 277.15 k
V1 = 0.5 L
Let's find the final volume (V2).
To solve for V2 apply Charles Law formula below:
The beam is refracted at 53.17-degree angle (asin 0.800414 = 53.17-degree). This problem can be solved by using the Snell's Law which described the refracted beam angle which traveled through a different media, in this case, through water and air. The formula of Snell's Law is stated as n1 sin θ1= n2 sin θ2. In this formula n1 is the refraction index of medium 1, θ1 is the normal light angle in the medium 1, n2 is the refraction index of medium 2, and θ2 is the normal light angle in the medium 2<span>. </span>
Answer:
V2 = 21.44cm^3
Explanation:
Given that: the initial volume of the bubble = 1.3 cm^3
Depth = h = 160m
Where P2 is the atmospheric pressure = Patm
P1 is the pressure at depth 'h'
Density of water = ρ = 10^3kg/m^3
Patm = 1.013×10^5 Pa.
Patm = 101300Pa
g = 9.81m/s^2
P1 = P2+ρgh
P1 = Patm +ρgh
P1 = 1.013×10^5+10^3×9.81×160.
P1 = 101300+1569600
P1 = 1670900 Pa
For an ideal gas law
PV =nRT
P1V1/P2V2 = 1
V2 = ( P1/P2)V1
V2 = (P1/Patm)V1
V2 = ( 1670900 /101300 Pa) × 1.3
V2 = 1670900/101300
V2 = 16.494×1.3
V2 = 21.44cm^3