How Does Entropy Change in an Isothermal Process for an Ideal Gas?

[Deathcrush]

Homework Statement

A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure
P 1 to pressure P 2 while loosing heat to the surroundings at temperature T in the amount of q. If the gas
constant of the gas is R, the entropy change of the gas ∆s during this process is
(a) ∆s =R ln(P 2 /P 1 ) (b) ∆s = R ln(P 2 /P 1 )- q/T (c) ∆s =R ln(P 1 /P 2 ) (d) ∆s =R ln(P 1 /P 2 )-q/T
(e) ∆s= 0

Homework Equations

PV=nRT (1)
u=q+w (2)
h=u+PV (3)
dw=-PdV (4)
ds=dq/T (5)

The Attempt at a Solution

first I considered h=0, since it is an isothermic process (I don't know if this is correct, this is one of my doubts)

I derived 2 and 3

du=dq+dw and du=-VdP-PdV

now, using 4 in this last derivative and comparing it with the first one, I found that dq=-VdP

and I used it in 5, also, I used 1 in V and integrated the whole thingS=R \int-dp/P) from state 1 to state two

I integrated from state two to one, and changed the negative sign

this would be something like
S=R Ln(P1/P2)
after using the properties of the logarithms

My biggest doubt is that of the enthalpy, if it is not 0, the whole thing is wrong. I chequed the answers in the book and mine is right, but I don't know if the process is ok, would anybody help me on this please?

 

[Mapes]

It may be of use to know that the enthalpy H of an ideal gas is H=C_P\Delta T+H_0, where C_P is the heat capacity, \Delta T is the temperature change, and H_0 is an arbitrary reference value.

 

[Andrew Mason]

Your have the right answer but it is simpler than you have done.

Since dT = 0, dU=nCvdT = 0. So dQ = PdV (Use the first law dQ = dU + dW where dW is the work done BY the gas = PdV)

dV/dP = d/dP(nRT/P) = -nRT/P^2

So:

\Delta S = \int ds = \int dQ/T = \frac{1}{T}\int_{P_1}^{P_2} PdV = -nR\int_{P_1}^{P_2}\frac{dP}{P}

AM

 

[Deathcrush]

It may be of use to know that the enthalpy H of an ideal gas is H=C_P\Delta T+H_0, where C_P is the heat capacity, \Delta T is the temperature change, and H_0 is an arbitrary reference value.

thanks a lot both of you, now, just another doubt, is H=C_P\Delta T+H_0 valid only for ideal gases? because I am working out another problem now, and I was planning to use the equation I got in this problem, but now it is not an ideal gas, it is air at about 25atm, is the equation valid?

 

[Mapes]

It is valid only for ideal gases. Depending on the nonideality, you may need to add a correction term (aka an enthalpy departure function).

 

[Andrew Mason]

thanks a lot both of you, now, just another doubt, is H=C_P\Delta T+H_0 valid only for ideal gases? because I am working out another problem now, and I was planning to use the equation I got in this problem, but now it is not an ideal gas, it is air at about 25atm, is the equation valid?

This is true only for constant pressure processes. It is simply saying that for constant pressure processes \Delta Q = H - H_0 = C_p\Delta T, ie. just a special case of applying the first law. In this case, pressure is not constant. Temperature is. So you just apply the first law: \Delta Q = \int PdV. There is positive heat flow into the gas in this case.

AM

 

[Andrew Mason]

If you want to determine the change in enthalpy in this process, start with the definition:

H = U + PV

This means that:

dH = dU + PdV + VdP

In this case, dU = 0 so dH = PdV + VdP:

\Delta H = \int_{V_i}^{V_f} PdV + \int_{P_i}^{P_f} VdP = nRT( \int_{V_i}^{V_f} <br /> \frac{dV}{V} + \int_{P_i}^{P_f} \frac{dP}{P}) = nRT\left(\ln{\frac{V_f}{V_i}} + \ln{\frac{P_f}{P_i}}\right)

But since T is constant, Vf/Vi = Pi/Pf, so

\Delta H = 0

AM

 

[Deathcrush]

No, I am actually trying to find the intermediate pressure for a 2 step compression so that the work needed is minimum