Archived Thermodynamics: Calculate the volume and temperature

skrat

1. Homework Statement
In an isolated chamber closed by a moving piston is an capsule with 0.5 l of air at 3 bar and 100 °C. Chamber is filled with 2 l of air at 1 bar and 20 °C. The same pressure and temperature surround the piston from the other side. Capsule breaks. What is the volume and temperature now? $\kappa =1.4$

2. Homework Equations

$\Delta U= \Delta W + \Delta Q = \Delta W + 0$ because of the isolation

$p_cV_c=\frac{m_c}{M}RT_c$ - c for capsule

$p_0V_0=\frac{m_0}{M}RT_0$ - 0 for gas before the capsule breaks

$p_0V=\frac{m_0+m_c}{M}RT$ - no index for gas after the capsule breaks

3. The Attempt at a Solution

I would be really happy if somebody could check what am I doing wrong because I just can't get the same result as written in the book...

$\Delta U=m_cc_v(T-T_c)+m_0c_v(T-T_0)=(m_c+m_0)c_vT-m_cc_vT_c-m_0c_vT_0$

$\Delta W=-p_0(V-(V_c+V_0))$

$(m_c+m_0)c_vT-m_cc_vT_c-m_0c_vT_0=-p_0(V-(V_c+V_0))$

Now for all the indexes: $m_iT_i=\frac{p_iV_iM}{R}$ . Than:

$\frac{M}{R}c_v(p_0V-p_1V_1-p_0V_0)=-p_0V+p_0(V_c+V_0)$

for $cv=\frac{R}{M(\kappa -1)}$

$\frac{1}{\kappa -1}(p_0V-p_1V_1-p_0V_0)+p_0V=p_0(V_c+V_0)$

$p_0V(\frac{1}{\kappa -1}+1)-\frac{1}{\kappa -1}(p_1V_1+p_0V_0)=p_0(V_c+V_0)$

$p_0V(\frac{\kappa }{\kappa -1}=p_0(V_c+V_0)+\frac{1}{\kappa -1}(p_1V_1+p_0V_0)$

$V=\frac{(\kappa -1)p_0(V_c+V_0)+(p_1V_1+p_0V_0)}{p_0\kappa }$

For data above, this gives me 3.2 l instead of 2.71 l.

http://web2.0calc.com/?q=(0.4*10**5*(0.0005+0.002)+3*10**5*0.0005+10**5*0.002)/(10**5*1.4

WHAT ON EARTH is wrong here? :/

• tkyoung75
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Chestermiller

Mentor
I get 3.21 liters also. It looks like they forgot to include the 0.5 liters for the capsule. So I confirm skrat's answer. I get 23 C for the final temperature.

Last edited:
• nawab leghari

Chestermiller

Mentor
Looks good. For what its worth, 0.5L at 3bar expanding to take up only 0.7L at 1bar is a bit of a stretch (pV = constant, ideal gas approximation).
This is a completely incorrect interpretation of what happens. The gas originally in the capsule has a temperature of 100 C, and its temperature changes to 23 C. So pV is not constant. Also, the capsule gas occupies the entire 3.2 liters in the final state, not 0.7 L. And finally, it's final partial pressure is not 1 Bar. It is only about 1/3 bar. The number of moles of gas originally in the capsule is about 0.043. Try pV=(3.2)(0.33)=nRT=(0.043)(0.082)(296) and see what you get.

Chestermiller

Mentor
If the capsule gas occupies the entire 3.2L then what happens to the other 2L that is injected (edit: into the chamber)?
The gases are intermingled in the final state, and molecules of both gases are present throughout the entire chamber.
If the final pressure is not 1 bar then why would the question mention that the pressure on the other side of the piston is one bar?
Who said that the final pressure is not 1 bar? I said that the final partial pressure of the gas that was originally in the capsule is 1/3 bar in the final state. Are you familiar with the concept of partial pressure?
I think you are both aware that the methodology is not strictly accurate
Who says? You?
but you don't have to say its "completely incorrect".
Yes, I do, because it really is completely incorrect.
It provides a perfectly valid ball park figure (upper limit) for the final volume.
No. I stand by what I said. Now, if you continue to pursue this incorrect avenue of discussion, I will have to issue you a misinformation warning.

"Thermodynamics: Calculate the volume and temperature"

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