Simple (Frustrating) Combined Gas Law Problem

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AI Thread Summary
The discussion revolves around a gas law problem involving a gas initially at 562L, 700 torr, and 213 degrees Celsius, which is transferred to a 25.3L tank at 35 degrees Celsius. The initial calculation using Celsius temperatures yielded an incorrect pressure of approximately 2600 torr, while the correct answer is 13 atm or 9880 torr. The error was identified as a misuse of Celsius instead of converting to Kelvin for accurate calculations. Once the temperatures were converted to Kelvin, the correct pressure was achieved. The participants discussed the importance of using the Kelvin scale for thermodynamic calculations.
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Homework Statement



If 562L of a gas is prepared at 700 torr and 213 degrees C, and it is then pumped into a 25.3L tank at 35 degrees C, what pressure will the tank have to withstand?

Homework Equations



\frac{V_{1}*P_{1}}{T_{1}}=\frac{V_{2}*P_{2}}{T_{2}}

Becomes:

\frac{T_{2}}{T_{1}}*\frac{V_{1}}{V_{2}}*P_{1}=P_{2}


The Attempt at a Solution



This is pretty straight forward. That's why I'm getting so frustrated. I just plug and play and I should get the right answer.

\frac{35degC}{213degC}*\frac{562L}{25.3L}*700torr=~2600torr or 3.421 atm

According to my professor, the answer is supposed to be 13atm or 9880torr.

I just don't understand how this approach is not working since I am only missing one variable. It seems perfect for the combined gas law. Thank you for your time.


DDnow

P.S. Sorry about the botched latex attempt. I tried :)
 
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Hi DDNow,

DDNow said:

The Attempt at a Solution



This is pretty straight forward. That's why I'm getting so frustrated. I just plug and play and I should get the right answer.

\frac{35degC}{213degC}*\frac{562L}{25.3L}*700torr=~2600torr or 3.421 atm

The Celsius temperature scale is not a thermodynamic temperature scale, and so cannot be used here. Try converting your temperatures to the Kelvin scale.
 
Hey! Thank you for that. Worked perfectly. I assumed since it was a temperature divided by a temperature that the ratio would be the same for C or K. Obviously, I was wrong. I really appreciate your time alphysicist!

Edit: I'm not sure if I need to mark this as resolved or whatnot. If someone wants to tell me if there is another step I would appreciate it.
 
Last edited:
DDNow said:
Hey! Thank you for that. Worked perfectly. I assumed since it was a temperature divided by a temperature that the ratio would be the same for C or K. Obviously, I was wrong. I really appreciate your time alphysicist!

Edit: I'm not sure if I need to mark this as resolved or whatnot. If someone wants to tell me if there is another step I would appreciate it.

Sure, glad to help! As far as C or K, temperature differences are the same for both scales, but as you found the ratios are different.

There used to be a way to mark the threads as solved, but I don't believe that function is available right now.
 

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