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 Dec 2nd 2018, 02:13 PM #1 Junior Member   Join Date: Dec 2018 Posts: 4 Thermodynamics Need a little help please. I've got a question asking work out thermal transfer and work done using Charles Law when final temp is 250k where P = ( v - 4 )^2. 5kg gas expands from 300 to 400mm^3. Cp = 1005j/kg K. Its the P = ( v - 4)^2 bit that is throwing me. Thanks for any help
 Dec 3rd 2018, 04:19 AM #2 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 277 Your question is a little odd. Do you have the original question? Can you please scan it and upload it or link it? topsquark likes this.
 Dec 3rd 2018, 03:33 PM #3 Junior Member   Join Date: Dec 2018 Posts: 4 5kg of a gas expands from 0.3to 0.4 m3 according to Charles law evaluate thermal energy transfer & work done for the gas. Cp =1005 J/kg K. final gas temp is 250K where P = ( v-4 )2. That’s squared not sure how to do a square symbol on here.
 Dec 5th 2018, 04:04 AM #4 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 277 Sorry for the late reply... I've been quite busy at work over the past few days. My thermodynamics is also a bit rusty because I haven't done any for a while, so feel free to point out any issues. I think there's also some issues with the question... You state that the expansion occurs due to Charle's law, which indicates that the pressure is held constant during the gas expansion so that: $\displaystyle \frac{V_1}{T_1} = \frac{V_2}{T_2}$ However, you also state that there is a relationship between pressure and volume of the form $\displaystyle P = (V - 4)^2$ This law is very odd because it seems to suggest that the pressure is negative for the expansion states when $\displaystyle V_1$ or $\displaystyle V_2$ is substituted into the law. Is what you typed exactly what the question states? If so, it is an awful question. If we continue under the assumption that Charles' law is valid, then: mass = $\displaystyle m = 5$ kg heat capacity = $\displaystyle 1005$ J/kg/K Iniitial state: $\displaystyle V_1 = 0.3 m^3$ $\displaystyle T_1 = ?$ $\displaystyle P_1 = ?$ Final state: $\displaystyle V_2 = 0.4 m^3$ $\displaystyle T_2 = 250$ K $\displaystyle P_2 = ?$ Under isobaric expansion, we can find $\displaystyle T_1$: $\displaystyle T_1 = \frac{V_1 T_2}{V_2} = \frac{0.3 \times 250}{0.4} = 187.5$ K. The thermal energy transfer to the gas can be calculated using the equation for sensible heat transfer: $\displaystyle \Delta Q = m c_p (T_2 - T_1)= 5 \times 1005 (250 - 187.5) = 314062.5$ J Finally, the work done by the gas is equal to the area underneath the pressure-volume curve for the expansion. If pressure is constant $\displaystyle W = \int - P dV = - P \Delta V$ But we don't know the pressure. In order to go further, we have to make assumptions about the kind of gas we have. A heat capacity of 1005 J/kg/K suggests that we have dry air, which has a molar mass of 0.029 kg/mol. If we assume that air behaves like an ideal gas, we have: $\displaystyle P = \frac{mRT_2}{V_2 M} = \frac{5 \times 8.314 \times 250}{0.4 \times 0.029} = 895905$ Pa This means that the work done is $\displaystyle W = - 895905 (0.4 - 0.3) = - 89590.5$ J The negative sign indicates that the work is done by the gas on the surroundings. Note: If we actually have a two-part question and the expansion is not isobaric, we can instead use stated pressure-relationship and we have $\displaystyle W = \int -P dV$ $\displaystyle = -\int_{0.3}^{0.4} (V-4)^2 dV$ $\displaystyle = -\int_{0.3}^{0.4} (V^2 - 8V + 16) dV$ $\displaystyle = -\left[ \frac{V^3}{3} - 4V^2 + 16V \right]_{0.3}^{0.4}$ $\displaystyle = -\left(0.0213 - 0.64 + 6.4 - 0.009 + 0.36 - 4.8\right)$ $\displaystyle = -1.3323$ J This result is really off to me. It could be that the question you have just wants to test your ability to perform integration, but it's inconsistent with the rest of the question and gives a result that is not realistic at all. topsquark likes this. Last edited by benit13; Dec 7th 2018 at 03:09 AM. Reason: Typos
 Dec 6th 2018, 02:22 PM #5 Junior Member   Join Date: Dec 2018 Posts: 4 Thanks mate Yes the question is very vague in describing what kind of process it goes through. I don't thinks its a genuine question trying to get an actual real world value just testing to see if you can use what you've learnt to get an answer. To answer your other question about a second part it also asks to evaluate the same for Boyles law. How do you only use one to get the answer for work done and thermal transfer? I thought you needed to combine them? Blimey you answer posts late at night or early morning depending which way you look at it. zzzz Thanks for the help
Dec 7th 2018, 03:23 AM   #6
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 Originally Posted by Bigron1 To answer your other question about a second part it also asks to evaluate the same for Boyles law. How do you only use one to get the answer for work done and thermal transfer? I thought you needed to combine them?
Boyle's law states that there is a relationship between pressure and volume when the temperature is held constant.

$\displaystyle P \propto \frac{1}{V}$

Therefore

$\displaystyle P_1 V_1 = P_2 V_2$

Our previous calculation for pressure uses $\displaystyle T_2$ and $\displaystyle V_2$, so if we say that the pressure there is $\displaystyle P_2$, we can calculate $\displaystyle P_1$

$\displaystyle P_1 = \frac{P_2 V_2}{V_1} = \frac{895905 \times 0.4}{0.3} = 1194540$Pa.

The thermal energy exchange is zero because the temperature doesn't change. We used the value T = 250 K.

The work done can be calculated using integration:

$\displaystyle W = \int -P dV$
$\displaystyle = -\frac{mRT}{M} \int_{0.3}^{0.4} \frac{1}{V} dV$
$\displaystyle = -\frac{mRT}{M} \left[ \ln V \right]_{0.3}^{0.4}$
$\displaystyle = -\frac{mRT}{M} \left( \ln 0.4 - \ln 0.3 \right)$
$\displaystyle = -\frac{mRT}{M} \frac{\ln 0.4}{\ln 0.3}$
$\displaystyle = -\frac{5 \times 8.314 \times 250}{0.029} \frac{\ln 0.4}{\ln 0.3}$
$\displaystyle = - 272733.6$ J

 Blimey you answer posts late at night or early morning depending which way you look at it. zzzz Thanks for the help
I am based on the UK, so the difference in time-zone probably makes my post appear as if they are at unusual hours. However, I can assure you I'm not up in the middle of the night solving physics problems!

 Dec 10th 2018, 03:10 PM #7 Junior Member   Join Date: Dec 2018 Posts: 4 Sorry for the slow reply Benit busy weekend. Hopefully final question for you. If I am supposed to integrate the P for both questions wouldn't I get the same answer for both questions? But then if I was supposed to do that then I would breaking the gas law what I am supposed to be using to work out the answer. Very confusing. Is that kind of what you are getting at? Cheers Bud
Dec 10th 2018, 03:40 PM   #8
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Posts: 277
 Originally Posted by Bigron1 Sorry for the slow reply Benit busy weekend. Hopefully final question for you. If I am supposed to integrate the P for both questions wouldn't I get the same answer for both questions? But then if I was supposed to do that then I would breaking the gas law what I am supposed to be using to work out the answer. Very confusing. Is that kind of what you are getting at? Cheers Bud
You won't get the same answer because one expansion is isobaric and the other is isothermal. The initial conditions required to achieve the final conditions under each case are different.

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