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We all know the basic understanding of the adiabatic phenomenon: ideal gas expands and cools, compresses and heats up. Now by expand, let take a quick look at a scenario: let assume there is a tank filled with nitrogen gas at 300°C pressurized to 4500 psi, 468 g of the gas is vented through a tube into a large balloon and the tank pressure drops to 120 psi in 10 seconds. The balloon reaches a pressure of 20 psi through the process. Based on adiabatics, to what temperature does the gas cools by expanding through the tube and into the balloon or does it even cools at all?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Feb 23, 2019 at 15:16
  • $\begingroup$ From original data: Pi= 307atm; Pf= 9.16atm; Ti= 573K; moles vented= 468/28= 1.671mole. From expansion eqn Tf= Ti(Pf/Pi)^.286 = 210K. find n (number of moles) from two simultaneous equations using PV=nRT: 1) 307*V=ni*(.08314)*(573); and 2) 9.16*V=(ni-1.671)*(.0831)*(210). Solve equations for n; ni= 1.818 moles and nf=.147 moles. V for tank: V= niRTi/Pi = .282L. Now assume the filling of balloon is NOT isobaric; i.e.;it is increasing from zero, so you can't use dU=PdV, but must use internal energy. Energy loss in the pressure vessel is - dU=niCvTi- nfCvTf = 210 L-atm. dU gain in balloon = $\endgroup$
    – anderh1
    Jan 31, 2022 at 20:48
  • $\begingroup$ [...] 210 atm-L=(1.671mol)(.208atm-L/mole-K)(T-balloon). T-balloon= 604K. Ideal gas law yields V-balloon= 61.7L. Gas in balloon gets hotter because it is being compressed. For reference, V at STP for 1.671 moles is 37.4L. $\endgroup$
    – SuperCiocia
    Feb 3, 2022 at 16:46

1 Answer 1

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$$T=(300+273)\left(\frac{134.7}{4514.7}\right)^{0.286}=210\ K=-63 C$$

ADDENDUM

Here are some calculations and analysis for your consideration based on ideal gas:

Initial pressure = 4500 psig = 307.1 atm.

Final pressure = 120 psig = 9.16 atm.

$$T_f=T_i\left(\frac{P_f}{P_i}\right)^{2/7}=210\ K=-63\ C$$ Calculation of initial moles $n_i$ in tank: $$n_i=\frac{500}{28}=17.857$$ Calculation of tank volume:$$V_T=\frac{n_iRT_i}{P_i}=\frac{(17.857)(0.08205)(573)}{307.1}=2.735\ liters$$ Calculation of final number of moles in tank: $$n_f=\frac{P_fV_T}{RT_f}=\frac{(2.735)(9.1633)}{(0.08205)(210)}=1.454\ moles=40.7\ grams$$ Application of the first law of thermodynamics to get the final temperature in the balloon:

Treating the mass of gas in the tank and the balloon as our closed system, and applying the first law of thermodynamics, we have: $$\Delta U=-\int_0^{V_f}{P_{ext}dV}$$where V is the volume of the balloon, and the change in internal energy of the gas is given by: $$\Delta U=n_fC_v(T_f-T_i)+(n_f-n_i)C_v(T-T_i)=(1.454)(2.5R)(210-573)+16.4(2.5R)(T-573)$$

We don't know the external pressure applied by the balloon rubber membrane to the gas because we don't know how the rubber behaves in terms of its stress strain behavior. But, we do know that the final pressure in the balloon is 20 psig. So, we rewrite the work term in our energy balance equation in a little different way:$$\int_0^{V_f}{P_{ext}dV}=\overline{P_{ext}}V_f=\frac{\overline{P_{ext}}}{P_{ext,f}}P_{ext,f}V_f$$where $$\overline{P_{ext}}=\frac{1}{V_f}\int_0^{V_f}{P_{ext}dV}$$We know that the ratio of the average external pressure to the final external pressure is going to be less than 1 because the pressure in the balloon is going to be increasing monotonically with time. So we can re-express the work as $$\int_0^{V_f}{P_{ext}dV}=kP_{ext,f}V_f$$with (0 < k < 1). From the ideal gas law, we also know that $$P_{ext,f}V_f=(n_i-n_f)RT$$So, combining previous equations, we have: $$n_fC_v(T_f-T_i)+(n_f-n_i)C_v(T-T_i)=-k(n_i-n_f)RT$$or equivalently: $$(1.454)(2.5)(210-573)+16.4(2.5)(T-573)=-k(16.4)T$$ Solving this for values over the allowable range of k, we obtain:

T = 603 K = 325 C for k = 0

T = 432 K = 159 C for k = 1

T = 506 K = 234 C for k = 0.5

Questions?

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Feb 23, 2019 at 15:16
  • $\begingroup$ So in your newly added analysis, the final values for Ts over the k interval (0,1) are the final temperature for what? And what condition does k represent in a practical sense. If they are the final temperature in the balloon, then where goes the adiabatic cooling. $\endgroup$
    – TechDroid
    Feb 24, 2019 at 13:15
  • $\begingroup$ For we have come this far, I think it's time I pour out my thoughts and understanding of the situation so far. Since the adiabatic cooling happens in the tank, I believe as temperature in the tank drops, temperature of gas leaving it also drops. Now let assume we track every 10 g of nitrogen leaving the tank, so definitely every 10 g is cooler than the last, and the last 10 is as cool as roughly ~-50°C. So averaging out the total change in temperature from 300°C to -50°C, we should be at a temperature low enough. $\endgroup$
    – TechDroid
    Feb 24, 2019 at 13:16
  • $\begingroup$ I'm not sure the red-kite guys on the site would be happy with further extending the comments. I don't know why my messages are not reaching your end, maybe try to refresh the page or check some settings or something. $\endgroup$
    – TechDroid
    Feb 24, 2019 at 13:31
  • $\begingroup$ The T's are the final temperatures in the balloon. The adiabatic cooling takes place in the tank. But the pressure in the balloon starts out at 0 psig, and ends up at 20 psig. So, some recompression of the gas occurs, and this recompression, along with the viscous heating in the valve, such that there is no temperature change in the valve (even though there is an additional pressure drop through the valve) , results in temperature recovery. If no work were done, the decrease in internal energy in the tank would actually be offset by an internal energy increase in the balloon (i..e., k = 0) $\endgroup$ Feb 24, 2019 at 14:06

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