At What Temperature in °F Will the Turkey Be Done?

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The discussion revolves around calculating the temperature at which a turkey is done using a pop-up thermometer with trapped air. The initial conditions include a pressure of 750 mmHg and a temperature of 10°C, while the pressure needed to pop the thermometer is 932 mmHg. Participants explore the Ideal Gas Law and combined gas law to find the final temperature, with calculations yielding results around 176°F and 174°F. There is confusion regarding the use of units and the application of gas laws, particularly concerning constant volume assumptions and conversions. Ultimately, the conversation highlights the importance of careful calculations and unit consistency in thermodynamic problems.
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Homework Statement


A pop-up turkey thermometer has 1.0 mL of air trapped beneath the red plastic indicator. When the thermometer was assembled, the temperature was 10 *C, and atmosphereic pressure was 750 mmHg. If the pressure required to break the seal and pop up the syringe is 932 mm Hg, at what temperature in *F will the turkey be done?


Homework Equations


PV=nrT
*C x 9/5 + 32 = *F
*C + 273 = *K
P_1V_1=P_2V_2
1 atm = 760 mmHg
P_1V_1/n_1T_1=P_2V_2/n_2T_2 (I'm not sure if I need this one or not)


The Attempt at a Solution



First, I started computing my starting variables.

P_1 = 750/760 = .987 atm
V_1= .001 L
T_1= 283 *K
n_1 = ?

P_2= 932/760 = 1.23 atm
V_2 = ?
T_2 = ?
n_2 = ?

To continue, I need to find n. So I used the Ideal Gas Law (no excessive pressures).

PV=nRT
PV/RT=n
(.987 atm)(.001 L)/(.0821 Lxatm/molxK)(283 *K) = 4.25 x 10^-5 mol air

I assign this value to both n_1 and n_2, because the mols of air shouldn't change.

When writing the combined gas law, I still need to find V_2 to find T_2. So I use Charle's Law.

P_1V_1/P_2 = V_2
(.987 atm)(.001 L)/(1.23 atm) = 8.02 x 10^-4

This is where I'm uncertain. I rearrange the combined gas law to look like this:

P_1V_1n_2/P_2V_2n_1T_1 = T_2

Plug in my numbers, I get 3.54 x 10^-3 *K. That's rediculous, considering the context of the problem. So I try to use the Ideal Gas Law, but only on the second state of gas

PV=nRT
manipulated to PV/nR=T

(1.23 atm)(.001 L)/(4.248 x 10^-5 mol)(.0821 L x atm/mol x K) = 353 *K. A much more reasonable answer. I convert to C, and get 80 *C. I think I've got it. Convert to *F, and I get 176 *F. But the computer says I'm wrong. Where did I mess up? Help please?
 
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It seems to be essentially a bomb problem with V and n constant. Try P1/T1 = P2/T2 using the original mmHg values, this is essentially your same calculation but with all the conversions it may be a matter of rounding that throws your answer off from what the computer has.
 
Last edited:
eli64 said:
It seems to be essentially a bomb calorimetry problem with V and n constant. Try P1/T1 = P2/T2 using the original mmHg values, this is essentially your same calculation but with all the conversions it may be a matter of rounding that throws your answer off from what the computer has.

Thank you for the input, but I'm thinking the combined gas law isn't the right direction. The P1/T1=P2/T2 yields -459 *F. I tried to put the answer in anyway, and nope. I also tried the calculations that yielded a reasonable answer while not eliminating my sig figs until I reached the very end, and still nothing. I'm all for blaming the computer for my problems, but I have a feeling the problem is me. . . any other suggestions or hints?
 
Not sure how you got the -459*F. If P1=750mmHg, T1=283K, P2=932mmHg then T2=351K or 78.7*C which is 174*F. P does not have to be in atm but T must be in K

If the volume is not constant, meaning that the expansion must include the new volume, then it seems data is missing or someone else may have other suggestions
 
Thanks. Must've just been a miscalculation on my part.

I thought it had to be in atm. Didn't realize you could use mmHg.
 
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