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# thermodynamics

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An ideal gas is connected in a closed cylinder in a moving piston the initial pressure is 1ATM and the start is 1 L. Constant pressure gas is heated up to double its volume then heated to constant volume to double the pressure and finally expands adiabatic until the temperature drops to its initial value.

un gas ideal esta conectado en un cilindro cerrado en un piston movil la presion inicial es 1atm y el inicio es 1L. se calienta el gas a presion constante hasta ducplicar su volumen despues se calienta a volmen constante hasta duplicar la presion y finalmente se expande adiabaticamente hasta que la temperatura deciende a su valor inical

dimcorak  Oct 11, 2017
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An ideal gas is connected in a closed cylinder in a moving piston the initial pressure is 1ATM and the start is 1 L. Constant pressure gas is heated up to double its volume then heated to constant volume to double the pressure and finally expands adiabatic until the temperature drops to its initial value.

un gas ideal esta conectado en un cilindro cerrado en un piston movil la presion inicial es 1atm y el inicio es 1L. se calienta el gas a presion constante hasta ducplicar su volumen despues se calienta a volmen constante hasta duplicar la presion y finalmente se expande adiabaticamente hasta que la temperatura deciende a su valor inical

Hallo dimcorak!

0. Ein ideales Gas ist in einem Zylinder mit einem beweglichen Kolben eingeschlossen.

Das Anfangsvolumen ist 1 Liter, der Anfangsdruck 1013 hPa.

1. Das Volumen wird isobar verdoppelt.

2. Der Druck wird isochor verdoppelt.

3. Die Temperatur sinkt bei adiabatischer Expansion auf die Anfangstemperatur.

(Only for my understanding.)

$$Pressure, temperature\ and\ volume\\ according\ to\ the\ three\ state\ changes.$$

0.

$$p_0=1013\ hPa\\ V_0= 1000\ cm^3$$

T is undetermined. I suggest:

$$T_0=20°C=293K$$

1. isobar

$$p_1=1013\ hPa\\ V_1= 2000\ cm^3$$

$$\frac{V_1}{T_1}=\frac{V_0}{T_0}\\ T_1=\frac{V_1\cdot T_0}{V_0}=\frac{2l\cdot293K}{1l}\\ \color{blue} T_1=586K$$

2. isochor

$$\color{blue}V_2=V_1=2000\ cm^3\\ \color{blue}p_2=2p_1=2026hPa\\ \frac{p_1}{T_1}=\frac{p_2}{T_2}\\ T_2=\frac{p_2T_1}{p_1}=\frac{2026\cdot 586K}{1013}\\ \color{blue}T_2=1172K$$

$$\kappa_{gas\ ideal}=1.4$$

$$\color{blue }T_3=T_0=293K$$

$$\frac{T_2}{T_0}=(\frac{V_2}{V_3})^{\kappa-1}\\ (\frac{T_0}{T_2})^{\frac{1}{\kappa-1}}=\frac{V_3}{V_2}\\ V_3=V_2(\frac{T_0}{T_2})^{\frac{1}{\kappa-1}}$$

$$V_3=2000cm^3(\frac{1172K}{293K})^{\frac{1}{1.4-1}}$$$$\frac{p_0}{p_2}=(\frac{V_2}{V_3})^{\kappa}$$

$$V_3=64000\ cm^3$$

$$\frac{p_3}{p_2}=(\frac{V_2}{V_3})^{\kappa}\\ p_3=p_2(\frac{V_2}{V_3})^{\kappa}\\ p_3=2026hPa(\frac{2000cm^3}{64000cm^3})^{1.4}\\$$

$$p_3=15.828\ hPa$$

greeting

!

asinus  Oct 11, 2017
edited by asinus  Oct 11, 2017
edited by asinus  Oct 12, 2017
#2
+7086
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Thank you for your "thank you".

!

asinus  Oct 12, 2017
#3
+31
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Thank you. Seriously, you were very helpful, you didn't understand the question.

dimcorak  Oct 18, 2017

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