The problem statement is as follows:

An insulated rigid vessel of volume $V$ initially contains $N_i$ moles of an ideal gas at temperature $T_i$. A pipe is now connected to the vessel and the same type of gas (but held at a different constant temperature $T_\text{in}$ in the pipe) is slowly pumped into the vessel until the total number of moles in the vessel becomes $N_f$. Obtain an explicit expression for the final temperature in the vessel in the form $T_f = f(T_i, T_\text{in}, N_f/N_i, \gamma)$ where $\gamma = C_p/C_v$, the ratio of molar isobaric heat capacity of the gas $C_p$ to the molar isochoric heat capacity of the gas $C_v$, both of which are independent of temperature. Assume there is negligible heat transfer between the gas in the tank and the vessel walls during the process.

My thoughts are that the energy it takes to heat the initial moles in the tank, $N_i$, to the final temperature, $T_f$, is equal to $(N_i)(C_v)(T_f-T_i)$. Would this be equal to the energy lost by the incoming gas $(N_f-N_i)(C_p)(T_\text{in}-T_f)$? Then all one would need to do is isolate $T_f$ in terms of the other variables. I am unsure if I'm missing some energy term and if my equating of $\Delta U$ of the preexisting gas to the $\Delta H$ of the incoming gas is correct.


1 Answer 1


It looks similar to this problem in this book ("FUNDAMENTALS OF ENGINEERING THERMODYNAMICS" by By E. RATHAKRISHNAN) on page 67 (uniform flow process):


You perform an energy balance on the control volume, and simplifying leads to: $$(N_f-N_i)*h_{in}=N_f u_f-N_i u_i$$

With: $$u_f = c_vT_f $$ $$ u_i = c_vT_i$$ $$h_{in} = c_pT_{in}$$

Therefore: $$T_f = (1-N_i/N_f).\gamma.T_{in}+ T_i.(N_i/N_f)$$

  • $\begingroup$ I saw this answer previously...but I'm a bit confused as how it is derived. I don't have access to that text. $\endgroup$ Jan 7, 2016 at 16:21
  • $\begingroup$ Follow the link I posted..it's in google books and it's free to view that part of the text. $\endgroup$
    – kitt91
    Jan 7, 2016 at 16:57
  • $\begingroup$ Sorry, I missed it when I initially followed the link. Thanks! $\endgroup$ Jan 7, 2016 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.