Simple (Frustrating) Combined Gas Law Problem

[DDNow]

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 :)

 

[alphysicist]

Hi DDNow,

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.

 

[DDNow]

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.

 

[alphysicist]

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.