Solve a Gas Law Problem: Find T in °C

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Discussion Overview

The discussion revolves around solving a gas law problem to find the temperature of an ideal gas sample in degrees Celsius, given its pressure, volume, and number of moles. Participants explore the application of the ideal gas law equation, PV=nRT, and address arithmetic errors in the calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a calculation using the ideal gas law and arrives at a temperature of -252 degrees Celsius, expressing confusion over the correctness of this result.
  • Another participant points out an arithmetic error but does not specify the nature of the error, suggesting that the correct temperature should be in the 200s.
  • A later reply emphasizes the importance of careful arithmetic and clarifies that the temperature calculated should be significantly higher than initially stated, with one participant calculating a temperature of 3104 K.
  • Participants discuss the need for showing individual steps in the calculations to identify where errors may have occurred.

Areas of Agreement / Disagreement

There is no consensus on the correct temperature calculation, as participants express differing views on the arithmetic involved and the resulting temperature. The discussion remains unresolved regarding the correct final temperature.

Contextual Notes

Participants have not fully clarified the steps taken in their calculations, leading to uncertainty about the arithmetic process and the final temperature result.

Jennifer_T
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Homework Statement


If 1.15 moles of an ideal gas has a pressure of 3.60 atm, and a volume of 81.37 L, what is the temperature of the sample in degrees Celsius?

Homework Equations


PV=nRT
p = pressure
v = volume
n = number of moles
r = 0.08206 L atm/(mol K)
T = ?

The Attempt at a Solution


So I did the problem by :
3.60 (81.37) = (1.15)(0.08206)(T)
I found T to be 20.9 degrees K. I converted the temperature from K to C by doing K = C + 273 so 20.9 = C + 273. So the temperate in C is -252 degrees. However, I am still getting this wrong in the online homework. Can anyone help me figure out mistake?
 
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Jennifer_T said:

Homework Statement


If 1.15 moles of an ideal gas has a pressure of 3.60 atm, and a volume of 81.37 L, what is the temperature of the sample in degrees Celsius?

Homework Equations


PV=nRT
p = pressure
v = volume
n = number of moles
r = 0.08206 L atm/(mol K)
T = ?

The Attempt at a Solution


So I did the problem by :
3.60 (81.37) = (1.15)(0.08206)(T)
I found T to be 20.9 degrees K. I converted the temperature from K to C by doing K = C + 273 so 20.9 = C + 273. So the temperate in C is -252 degrees. However, I am still getting this wrong in the online homework. Can anyone help me figure out mistake?
You did the arithmetic incorrectly. That's all.

Chet
 
Chestermiller said:
You did the arithmetic incorrectly. That's all.

Chet

I have done the problem multiple times and am unsure of where my arithmetic problem is.
 
Jennifer_T said:
I have done the problem multiple times and am unsure of where my arithmetic problem is.
Well, your equation should give the correct temperature, which is in the 200's. I guess the only what we can help you further is if you show the individual steps in your arithmetic. What do you get for the product of PV? When you divided that by R what do you get? When you divide that by the number of moles, what do you get?

Chet
 
Chestermiller said:
Well, your equation should give the correct temperature, which is in the 200's. I guess the only what we can help you further is if you show the individual steps in your arithmetic. What do you get for the product of PV? When you divided that by R what do you get? When you divide that by the number of moles, what do you get?

Chet

So I did (3.60)(81.37)/(1.15)(0.08206) into my calculator in order to get T in Kelvins.
 
Jennifer_T said:
So I did (3.60)(81.37)/(1.15)(0.08206) into my calculator in order to get T in Kelvins.
Yes. That's correct. When I did it on my calculator, I got 3104K. That 0.08206 is in the denominator, not in the numerator. You need to be careful with your parenthesis.

Chet
 
I guess the temperature wasn't in the 200s after all. The surface of the sun is about 5700 K, so this gas sample is quite the hot potato.
 
SteamKing said:
I guess the temperature wasn't in the 200s after all. The surface of the sun is about 5700 K, so this gas sample is quite the hot potato.
Ha! I noticed that. That's what I get for trying to do it in my head.

Chet
 

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