Calculating Isothermal Expansion Heat, Work, Internal Energy and Entropy

[ipitydatfu]

i was given this test problem a couple of days ago, that ha2 parts:
1 mol of ideal
a) isothermal reversible at a Ti, Pi = 3 atm, Pf= 1 atm

b)irreversible isothermal expansion under constant pressure (1 atm) .

find the heat/work/internal energy, delta(H), entropy for both.


attempt:


part a is easy, that was well covered over the lecture/internets/various physics book

but for part b, i wasnt so sure

for my answers i put something in the lines of:

U = q + w, deltaU = 0
therefore
q = -w
H = 0 (not too sure)

i guess since its under constant pressure
q = -w, and w = -p (delta v)
q = (1 atm)(Vf - Vi) (volumes were found from ideal gas law)

and Entropy
S = Ssys + Ssur
S = q/t + Ssur

how do i get the entropy surroundings? and how far off am i?

edit: bump no love for this thread?

 

[lippyka]

Answer:For part a (isothermal reversible): U = 0, q = w, H = -nRT ln(Pf/Pi), S = nR ln(Vf/Vi)For part b (irreversible isothermal expansion under constant pressure): U = 0, q = -w, H = 0, S = q/T + SsurrFor the entropy of the surroundings, you can use the fact that the entropy of the surroundings is unchanged. That is, Ssurr = 0.

 

[kerimek]

Hello,

I would like to provide some feedback on your attempt at solving the problem.

For part a, it is correct to say that the internal energy change, deltaU, is equal to the heat, q, added to the system minus the work, w, done by the system. Since the process is isothermal, the internal energy change is zero, so q = w.

To calculate the heat, we can use the ideal gas law:
PV = nRT
where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

Since the process is isothermal, the temperature remains constant at Ti = Tf. We can rearrange the ideal gas law to solve for Vf:
Vf = (nRTi)/Pf
Substituting the values given, we get:
Vf = (1 mol)(0.08206 L*atm/mol*K)(Ti)/(Pf)
Vf = (1 mol)(0.08206 L*atm/mol*K)(273 K)/(1 atm)
Vf = 22.41 L

So the work done by the system is:
w = -P(Vf - Vi)
w = -(1 atm)(22.41 L - 22.41 L)
w = 0 J

Therefore, the heat added to the system is also 0 J.

For part b, you are correct in saying that q = -w. However, the work done by the system is not just the change in volume, but also the change in pressure. So we can calculate the work as:
w = -P(deltaV) - P(deltaP)
where deltaV is the change in volume and deltaP is the change in pressure.

Since the process is isothermal, the temperature remains constant, so the ideal gas law can be used to find the change in volume:
deltaV = (nRT)/Pf - (nRT)/Pi
deltaV = (1 mol)(0.08206 L*atm/mol*K)(Ti)/(Pf) - (1 mol)(0.08206 L*atm/mol*K)(Ti)/(Pi)
deltaV = (1 mol)(0.08206 L*atm/mol*K)(273 K)/(1 atm) - (1 mol)(0.08206 L*atm/mol*K)(273 K)/(3 atm)
deltaV =