• Support PF! Buy your school textbooks, materials and every day products Here!

Problem on thermodynamics first law again

  • Thread starter ky92
  • Start date
  • #1
14
0

Homework Statement


Consider a system consisting of 2.0 mol CO2 (assumed to be a perfect gas) at 298K confined to a cylinder of cross-section 10cm^2 at 10 atm. The gas is allowed to expand adiabatically and reversibly against a constant pressure of 1 atm. Calculate W, Q, ΔU, ΔH, and ΔT when the piston has moved 200cm.

(Ans: W=-1.6KJ, Q=0, ΔU=-1.6KJ, ΔT=-28, ΔH=-2.1KJ)

Homework Equations


The thermodynamics equations


The Attempt at a Solution


as it is adiabatic, Q=0.
ΔV=0.001m(0.2m)=2x10^-4 m3
W=-PexΔV=-101325(2x10^-4)=-20.27 J
ΔU=W+Q=-20.27 J

and i found that my W and ΔU are wrong
what's wrong with my answer?

should Pgas=Pex if the system is in equilibrium?
i just found that it isn't
but in my book :
"To achieve reversible expansion we must match Pex to P at each state: dw=-Pex dV=-P dV"

i am so confused.
please help!
 

Answers and Replies

  • #2
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
682
Calculate the work done by the gas on the piston:

[tex]W_{\text{on piston}} = \int_0^{l_f} P*A \,dl[/tex]

The work done on the gas is just the additive inverse of the work done by the gas. The piston moves 200 cm, given. The area is a constant, so you essentially need to integrate P*dl. The process is adiabatic. What is the relation between pressure and volume for an adiabatic process? (Hint: You need to use the fact that the gas is CO2.)
 
  • #3
Andrew Mason
Science Advisor
Homework Helper
7,562
325

Homework Statement


Consider a system consisting of 2.0 mol CO2 (assumed to be a perfect gas) at 298K confined to a cylinder of cross-section 10cm^2 at 10 atm. The gas is allowed to expand adiabatically and reversibly against a constant pressure of 1 atm. Calculate W, Q, ΔU, ΔH, and ΔT when the piston has moved 200cm.

(Ans: W=-1.6KJ, Q=0, ΔU=-1.6KJ, ΔT=-28, ΔH=-2.1KJ)
You make a good point. The problem is with the question. The gas can expand adiabatically but not reversibly. In order for it to be a reversible expansion, there has to be an infinitessimal pressure difference.

In this case there is a significant pressure difference. So what happens here is that some of the energy in the gas does work on the atmosphere and some does work on the piston. The work done on the piston means the piston acquires kinetic energy (or a combination of kinetic and gravitional potential energy if it is not horizontal). So it is a dynamic rather than a quasistatic or reversible process.

In order to calculate that the work that the gas does on the atmosphere and piston, use the internal pressure of the gas. As DH says, you have to use the relationship between P and V in an adiabatic process (adiabatic condition). Express P as a function of V and then as a function of A and length.

AM
 
  • #4
14
0
for an adiabatic process, Q=0,
ΔU=w+0=w
w=ΔU=CvΔT...

that's all i can think of

i have no idea lol
 
  • #5
14
0
You make a good point. The problem is with the question. The gas can expand adiabatically but not reversibly. In order for it to be a reversible expansion, there has to be an infinitessimal pressure difference.

In this case there is a significant pressure difference. So what happens here is that some of the energy in the gas does work on the atmosphere and some does work on the piston. The work done on the piston means the piston acquires kinetic energy (or a combination of kinetic and gravitional potential energy if it is not horizontal). So it is a dynamic rather than a quasistatic or reversible process.

In order to calculate that the work that the gas does on the atmosphere and piston, use the internal pressure of the gas. As DH says, you have to use the relationship between P and V in an adiabatic process (adiabatic condition). Express P as a function of V and then as a function of A and length.

AM
thank you!
let me think about it...
 
  • #6
Andrew Mason
Science Advisor
Homework Helper
7,562
325
for an adiabatic process, Q=0,
ΔU=w+0=w
w=ΔU=CvΔT...

that's all i can think of

i have no idea lol
But you also know that

[tex]PV^{\gamma} = P_0V_0^{\gamma} = K[/tex]

Since PV = nRT, P = nRT/V, so

[tex]TV^{\gamma-1} = K/nR[/tex]

which is also constant, ie [itex]TV^{\gamma-1} = T_0V_0^{\gamma-1} [/itex]

So work out the change in T after an adiabatic expansion of 200 cm (you have to first determine the initial volume and the expanded volume) using this expression. (What is the [itex]\gamma[/itex] for CO2?)

AM
 

Related Threads for: Problem on thermodynamics first law again

  • Last Post
Replies
2
Views
5K
Replies
9
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
1
Views
803
Replies
1
Views
2K
Replies
3
Views
1K
Replies
16
Views
954
Replies
1
Views
3K
Top