A closed system containing a gas is to undergo a reversible process

[MEM33]

Homework Statement

A closed system containing a gas is to undergo a reversible process from an intial specific volume of 2 ft^3/lbm and a intial pressure of 100 psia. The final pressure is 500 psia. Compute the work done per unit mass.

a. Pv= constant
b. pv^-2 = constant

Homework Equations

(1) p2/p1 = (v1/v2)^k

w = ∫Pdv

The Attempt at a Solution

Using equation (1) I solved for v2, w = ∫Pdv, but p is not constant, so i don't know how to find the work. Maybe step one is wrong, but I have the final answer = -59.6 Btu/lbm

Thanks!

 

[KataKoniK]

Hi there,

To solve this problem, we can use the first law of thermodynamics for a closed system, which states that the change in internal energy of the system is equal to the heat added to the system minus the work done by the system. In this case, we are only interested in the work done, so we can ignore the heat term.

From the ideal gas law, we know that for a reversible process, PV = constant. Therefore, we can use this relation to solve for the final specific volume, v2, in terms of the initial specific volume, v1, and the initial and final pressures, p1 and p2:

p1v1 = p2v2
v2 = (p1/p2)v1

Now, we can use this expression for v2 in our work equation, w = ∫Pdv, to solve for the work done per unit mass:

w = ∫p1v1^2/(p1/p2)^2 dv
w = ∫(p1/p2)^2v1^2 dv
w = (p1/p2)^2v1^2 ∫dv
w = (p1/p2)^2v1^2 (v2 - v1)

Substituting in our expression for v2, we get:

w = (p1/p2)^2v1^2 [(p1/p2)v1 - v1]
w = (p1/p2)^3v1^3 - (p1/p2)^2v1^2

Finally, we can plug in the given values for p1, p2, and v1 to solve for the work done per unit mass:

w = (100/500)^3(2)^3 - (100/500)^2(2)^2
w = (-0.008) - (0.016)
w = -0.024 ft^3-lbf/lbm

Converting to Btu/lbm, we get:

w = (-0.024 ft^3-lbf/lbm)(778.169 ft-lbf/Btu)
w = -18.7 Btu/lbm

So, the final answer is actually -18.7 Btu/lbm, not -59.6 Btu/lbm as stated in the forum post.

I hope this helps! Let me know if you have any further questions.