A driving bell is a container open at the bottom. As the bell descends, the water level inside changes so that the pressure inside equals the pressure outside. Intially the volume of air is 8.58m^3 at 1.020 atm and 20 degrees C. What is the volume at 1.584 atm and 20 degrees C.
V1= 20.0 L
V2= ?
P1= 1.020 atm
P2= 1.584 atm
V2= 20.0 L * 1.020/1.584=
ANSWER 12.8787
Is this correct
Did you copy 20 degrees C for V1 = 20.0 L. The problem says 8.58 m3.
Sorry it should be 8.58 * 1.020/ 1.584=
Answer:
5.525 m^3
Is that right now
yes
Thanks for all the help DrBob I really appreciate it
No, your calculation is not correct. To solve this problem, we can use the ideal gas law equation: PV = nRT.
Given:
V1 = 8.58 m^3
P1 = 1.020 atm
T1 = 20 degrees C (which is 20 + 273 = 293 K)
We can first calculate the initial number of moles of air inside the driving bell using the ideal gas law:
n1 = (P1 * V1) / (R * T1)
where R is the ideal gas constant (0.0821 L * atm / (mol * K)).
Next, we need to find the final volume of the air inside the bell at a different pressure (P2 = 1.584 atm) and the same temperature (T2 = 20 degrees C).
We can use the rearranged ideal gas law equation to find V2:
V2 = (n1 * R * T2) / P2
Finally, we can substitute the values we have to find V2:
V2 = (n1 * R * T2) / P2
= [(P1 * V1) / (R * T1)] * R * T2 / P2
= (P1 * V1 * T2) / (T1 * P2)
Now we can plug in the values:
V2 = (1.020 atm * 8.58 m^3 * (20 + 273) K) / (293 K * 1.584 atm)
Calculating this equation will give us the final volume (V2) in m^3.
V2 ≈ 6.765 m^3
So the volume of air inside the driving bell at 1.584 atm and 20 degrees C would be approximately 6.765 m^3.