Ideal gas adiabatic/isovolumetric help

  • Thread starter Thread starter scholio
  • Start date Start date
  • Tags Tags
    Gas Ideal gas
AI Thread Summary
The discussion revolves around calculating the change in entropy for an ideal gas undergoing an adiabatic expansion followed by an isobaric compression. The initial conditions include two moles of gas with specified pressure, heat capacities, and gamma value. Key points include the realization that the adiabatic process has no heat transfer (Q=0), resulting in no change in entropy during that phase. The second process is correctly identified as isobaric, requiring the use of Cp instead of Cv, and necessitating the calculation of temperatures post-expansion to determine entropy change. The need for a natural logarithm formula for entropy calculation is emphasized, highlighting the importance of accurately determining pressure and temperature throughout the processes.
scholio
Messages
159
Reaction score
0

Homework Statement


two moles of an ideal gas is initially at P = 2*20^5 Pa expands adiabaticaly to four times its original volume. it is then compressed at constant pressure to its original volume. what is the change in entropy of the gas

Cp = 20.78 joules/mole-deg
Cv = 12.47 joules/mole-deg
gamma = 5/3



Homework Equations



adiabatic expansion --> PV^gamma = constant, Q = 0
isovolumetric compression --> nCv(deltaT)
entropy deltaS= deltaQ/T

The Attempt at a Solution



i am not sure how to use the adiabatic equation because although P and V is given as well as gamma, i can solve for 'constant' but where does constant come into play?

since for an adiabat Q = 0, do i just take it as zero and move on or do i actually need to solve for something.

as for the isovolumetric compression, moles is given, Cv is given but not deltaT so i solved the Pv=nRT eq for T and subbed it in, i then assumed P and V to be constants as stated in the problem?? and solved for Q getting 1.5 joules

when i solved for entropy i did (compression - expansion, so isovolumetric minus adibat = i subbed in the equation from the isovolumetric equation in and canceled out the T's and was able to get deltaS = 24.94 joules per kelvin

did i take the correct approach? help appreciated...
 
Physics news on Phys.org
As you have recognized, the heat transferred in an adiabatic process is zero. Therefore, the ratio of Q/T is zero, and there is no change in entropy.

And the second process is at constant pressure (isobaric), not constant volume, so use Cp in the second equation.

You need to find the two temperatures, and for that you need to find the pressure after the adiabatic expansion. That's where the gamma equation comes in. (You said volume is given but you did not state what initial volume is). Once you have the pressure after the expansion, you can get the temperature simply using PV=nRT

Then find the delta T for the isobaric compression.

And to find the delta S, you need a slightly different formula, since T is not constant (look for a natural log formula).
 
hi thanks, just to clarify the problem does not specify an actual volume just the amount at which it decreases/increased
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top