At 9.20 cm cubed air bubble forms in a deep late at a depth where the temperature and pressure are 6 degress Celcius and 4.50 atm, respectively. Assuming that the amount of air in the bubble has not changed, calculate its new volume at STP.
Use the combined gas law for this:
P1V1/T1 = P2V2/T2
change the temps to Kelvins.
With anonymous answering anonymous I'll just agree that p1v1/t1 = p2v2/t2 is the formula to use and remember to change T to Kelvin
C _ 273 = K.
To solve for the new volume of the air bubble at STP (Standard Temperature and Pressure), we can use the combined gas law equation.
The combined gas law equation is:
P1V1/T1 = P2V2/T2
Where P1, V1, and T1 represent the initial pressure, volume, and temperature, respectively. P2, V2, and T2 represent the final pressure, volume, and temperature, respectively.
First, let's convert the initial temperature from Celsius to Kelvin:
T1 = 6°C + 273 = 279 K
Next, let's use the given information to assign values:
P1 = 4.50 atm
V1 = 9.20 cm³
T1 = 279 K
We also know that at STP:
P2 = 1 atm (since STP is defined as 1 atm)
T2 = 273 K (since STP temperature is defined as 0°C, which is equivalent to 273 K)
Now we can substitute the values into the combined gas law equation:
P1V1/T1 = P2V2/T2
4.50 atm * 9.20 cm³ / 279 K = 1 atm * V2 / 273 K
Let's solve for V2 by rearranging the equation:
V2 = (4.50 atm * 9.20 cm³ * 273 K) / (279 K * 1 atm)
Calculating this expression, we find:
V2 ≈ 38.60 cm³
Therefore, the new volume of the air bubble at STP is approximately 38.60 cm³.