Thermodynamics equilibrium problem

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SUMMARY

The forum discussion centers on a thermodynamics problem involving two rigid tanks containing nitrogen (N2) and oxygen (O2) gases. The initial conditions are 0.2 m³ of N2 at 350 K and 100 kPa in tank A, and 0.5 m³ of O2 at 500 K and 250 kPa in tank B. After mixing, the expected temperature is 474 K, but a participant calculated 497.061 K due to incorrect mass calculations. The correct approach involves using moles instead of masses and applying the first law of thermodynamics, leading to the conclusion that the heat lost during mixing can be modeled as Q = ΔU, where ΔU is the change in internal energy.

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  • Understanding of the Ideal Gas Law and its application (PV=nRT)
  • Knowledge of specific heat capacities and their role in thermodynamic calculations
  • Familiarity with the first law of thermodynamics (Q = ΔU)
  • Ability to perform calculations involving moles and mass conversions
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  • Study the concept of specific heat capacities for different gases
  • Explore the first law of thermodynamics in detail, focusing on adiabatic processes
  • Investigate the differences between using mass and moles in thermodynamic calculations
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This discussion is beneficial for students and professionals in mechanical engineering, chemical engineering, and anyone studying thermodynamics, particularly those dealing with gas mixtures and energy transfer in closed systems.

integ8me
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Please help me solve this...

Two rigid tanks are connected by a valve. Initially, tank A contains 0.2m3 of n2 at 350k and 100Kpa. tank B contains 0.5m3 O2 at 500k and 250Kpa. The valve between the tanks is open and the two gases are allowed to mix. Assuming constant specific heats at the given temp find the temp of the gases immediately after mixing (474k) and the amount of heat lost if the tanks are allowed to sit and reach equilibrium with the surroundings at 25c (145.8KJ).

I made the assumptions that if the tanks are the control volume initially then Q=0, KE=0, PE=0.
Ma=PV/RT = (100*0.2)/(0.2968*350)=0.19253kg
Mb=PV/RT = (250*0.5)/(.02598*500)=9.62279kg

Mmix = Ma +Mb
Tmix=(Ma/Mmix)*(Ta)+(Mb/Mmix)*(Tb)
I keep getting Tmix=497.061K
I'm supposed to get 474K, am I missing something obvious?

I included boundary work for the second part of the problem and cannot get the correct answer either. Please help if you can
 
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integ8me said:
Please help me solve this...

Two rigid tanks are connected by a valve. Initially, tank A contains 0.2m3 of n2 at 350k and 100Kpa. tank B contains 0.5m3 O2 at 500k and 250Kpa. The valve between the tanks is open and the two gases are allowed to mix. Assuming constant specific heats at the given temp find the temp of the gases immediately after mixing (474k) and the amount of heat lost if the tanks are allowed to sit and reach equilibrium with the surroundings at 25c (145.8KJ).

I made the assumptions that if the tanks are the control volume initially then Q=0, KE=0, PE=0.
Ma=PV/RT = (100*0.2)/(0.2968*350)=0.19253kg
Mb=PV/RT = (250*0.5)/(.02598*500)=9.62279kg

Mmix = Ma +Mb
Tmix=(Ma/Mmix)*(Ta)+(Mb/Mmix)*(Tb)
I keep getting Tmix=497.061K
I'm supposed to get 474K, am I missing something obvious?

I included boundary work for the second part of the problem and cannot get the correct answer either. Please help if you can

Re-check your calculation of the amount of oxygen initially present in tank B. You've slipped a decimal point somewhere.
 
integ8me said:
Please help me solve this...

Two rigid tanks are connected by a valve. Initially, tank A contains 0.2m3 of n2 at 350k and 100Kpa. tank B contains 0.5m3 O2 at 500k and 250Kpa. The valve between the tanks is open and the two gases are allowed to mix. Assuming constant specific heats at the given temp find the temp of the gases immediately after mixing (474k) and the amount of heat lost if the tanks are allowed to sit and reach equilibrium with the surroundings at 25c (145.8KJ).

I made the assumptions that if the tanks are the control volume initially then Q=0, KE=0, PE=0.
Ma=PV/RT = (100*0.2)/(0.2968*350)=0.19253kg
Mb=PV/RT = (250*0.5)/(.02598*500)=9.62279kg
Mmix = Ma +Mb
Tmix=(Ma/Mmix)*(Ta)+(Mb/Mmix)*(Tb)

The last formula is wrong. Use the moles of the gases, instead of their masses.
 
Hm the equation for Tmix seems to be derived from Q (lost by oxygen)=Q (absorbed by Nitrogen). Is this the correct way to model this problem? We have gases that are mixing up not that exchange Q through a separation interface...
 
Delta² said:
Hm the equation for Tmix seems to be derived from Q (lost by oxygen)=Q (absorbed by Nitrogen). Is this the correct way to model this problem? We have gases that are mixing up not that exchange Q through a separation interface...
Since there is no work done on or by the surroundings in this mixing, ##Q = \Delta U = n_aC_v\Delta T_a + n_bC_v\Delta T_b## (on the assumption that they behave as ideal gases). And, since it is adiabatic, Q = 0, so ##n_aC_v\Delta T_a = - n_bC_v\Delta T_b##. So it is really just a matter of finding n and initial T for each gas .

AM
 
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integ8me said:
Please help me solve this...I made the assumptions that if the tanks are the control volume initially then Q=0, KE=0, PE=0.
Ma=PV/RT = (100*0.2)/(0.2968*350)=0.19253kg
Mb=PV/RT = (250*0.5)/(.02598*500)=9.62279kg

I'm supposed to get 474K, am I missing something obvious?
what value for R do you use, shouldn't you use R=8,314. All the units are in SI (mind pressure is in KPa so you should multiply x 1000).
I included boundary work for the second part of the problem and cannot get the correct answer either. Please help if you can
There is no work done, use Q=\Delta U_{O_2}+\Delta U_{N_2}=(n_{O_2}+n_{N_2})C_v\Delta T, T_i=474K, T_f=298K)

Ok , i see now , you used R_{specific} and calculated masses , however you should use the moles of the gases not their masses as ehild noticed, for both questions.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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