Physics Help Forum Thermodynamics - Temperature change of argon gas

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 Dec 3rd 2017, 02:06 PM #1 Junior Member   Join Date: Dec 2017 Posts: 1 Thermodynamics - Temperature change of argon gas the temperature of n = 19 mol of argon gas is increased from T1 = 21 oC by Q = 4.4 kJ heat transfer, while the gas pressure is kept constant. What is the new gas temperature in Celsius degrees? Q=n*Cv*change in Temp as its a monoatomic gas I think this means that the equation is just Q=Cp*change in temp. But I'm not really sure. The Cv of Argon is 0.3122, with R being 0.2081 and Cp being 0.5203. My last attempts have come up with the final temp as 44.9 - where I had used the mass of the object as 749. I also have got 30.88and 29.4 by subbing the values into the unknown. I keep getting the answer marked as wrong and I'm not sure where I'm going wrong
 Dec 5th 2017, 04:18 AM #2 Member   Join Date: Oct 2017 Location: Glasgow Posts: 32 I think there are several mistakes somewhere, so let's start from the beginning. Yes, the formula to use here is $\displaystyle Q = mc_p\Delta T$ (Use cp because the question states that the heat transfer occurs at constant pressure). You can obtain a formula describing what you want by rearranging it to make the final temperature the subject: $\displaystyle Q = mc_p\Delta T$ $\displaystyle Q = mc_p (T_2 - T_1)$ $\displaystyle T_2 - T_1 = \frac{Q}{mc_p}$ $\displaystyle T_2 = T_1 + \frac{Q}{mc_p}$ where $\displaystyle T_2$ is the final temperature and $\displaystyle T_1$ is the initial temperature. Now... you have 19 moles of Argon gas. 1 mol of Argon gas has a mass of approximately 40g (because of the mass number on the periodic table), so 19 moles of Argon gas has a mass of $\displaystyle 19 \times 0.04 = 0.76$ kg. $\displaystyle T_2 = 21 + \frac{4400}{0.76\times 520} = 21 + 11.13 = 32.13$ degrees Celsius. Last edited by benit13; Dec 5th 2017 at 04:26 AM.
 Dec 5th 2017, 06:37 AM #3 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 688 If the gas is heated at constant pressure, it must expand. So it must do work against the surroundings. So the heat added must equal the gain in internal energy of the gas to the new temperature plus the work done. IOW not all the heat added goes to increasing the temperature. benit13 likes this.
Dec 5th 2017, 06:59 AM   #4
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 Originally Posted by studiot If the gas is heated at constant pressure, it must expand. So it must do work against the surroundings. So the heat added must equal the gain in internal energy of the gas to the new temperature plus the work done. IOW not all the heat added goes to increasing the temperature.
Yes, that's correct. That's why you must use $\displaystyle c_p$ in the formula for the heat transfer in the problem outlined above to get the correct answer. If you use $\displaystyle c_p$ in the heat transfer formula, you get the correct answer because the work done by the gas is factored out of the problem.

Note that if you know $\displaystyle c_v$, you can work out the change in temperature that would have occurred if the heat transfer was made instead at constant volume and therefore with no energy going into work done:

$\displaystyle T_2 = T_1 + \frac{Q}{mc_v} = 21 + \frac{4400}{0.76 \times 312} = 39.56$ degrees Celsius.

It's higher

 Dec 12th 2017, 10:12 AM #5 Junior Member   Join Date: Dec 2017 Location: Johnson City Posts: 2 Dont forget to use Cp in your calculations and you will get the final result 39,56 degrees Celsius. This is the final result that you need. Also if you still feel unsure and need extra explanation you may ask a question at https://studydaddy.com/physics-homework-help where you can find some additional information.
Dec 13th 2017, 03:28 AM   #6
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 Originally Posted by Knowledgesearcher Dont forget to use Cp in your calculations and you will get the final result 39,56 degrees Celsius. This is the final result that you need.
No. The answer is 32.13 degrees Celsius. That's the result obtained (using cp) in my first reply in this thread.

Last edited by benit13; Dec 13th 2017 at 03:39 AM.

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