A 16.5 cm3 bubble under the ocean at 2.1 atm and 12.5oC surfaces to 1.0 atm and 22.1oC. What is the new volume (in cm3)?
i used (p1)(V1)/T1=(P2)(V2)/T2
(2.1)(.0165)/285.65=(1)(V2)/295.25
THE ANSWER I GOT IS .0335
i changed 16.5 to liters which is why its .0165
is this correct?
Yes, your calculation is correct. You used the ideal gas law equation, \(PV/T = P_2V_2/T_2\), to solve for the new volume. Here's how you derived the answer:
1. Convert the initial volume from cm^3 to liters: 16.5 cm^3 = 0.0165 L.
2. Substitute the given values into the equation:
\( (2.1\, \mathrm{atm}) \times (0.0165\, \mathrm{L}) / (285.65\, \mathrm{K}) = (1.0\, \mathrm{atm}) \times (V_2) / (295.25\, \mathrm{K}) \).
3. Rearrange the equation to solve for V2:
\( V_2 = (2.1\, \mathrm{atm}) \times (0.0165\, \mathrm{L}) \times (295.25\, \mathrm{K}) / (1.0\, \mathrm{atm}) / (285.65\, \mathrm{K}) \).
4. Calculate the expression:
\( V_2 = 0.0335\, \mathrm{L} \).
So, the new volume is indeed 0.0335 L, or equivalently, 33.5 cm^3.