How Is dA Calculated for an Isothermal Expansion of an Ideal Gas?

[grandpa2390]

Homework Statement

Use the fundamental eq for the helmholtz eq to find dA for 1 mole of ideal gas as it expands isothermally from 5L to 15L at 298 Kelvin

Homework Equations

dA = -PdV - SdT
PV = nRT

The Attempt at a Solution

I tried to solve this the same way I did the problem on the previous page to calculate the isothermal compressibility, but it is not working out.

I know that dT is going to equal zero, so dA = -PdV but when I try to calculate P, I am not sure how factor in the change in V. I think I remember doing this at the beginning of the semester, but I can't remember. or find my notes. Any hints?

 

[MexChemE]

You need to express P as a function of V using the Ideal Gas law, and then integrate the equation from state 1 to state 2 (volume 1 to volume 2).

 

[grandpa2390]

You need P as a function of V using the Ideal Gas law, and then integrate the equation from state 1 to state 2.

You need to express P as a function of V using the Ideal Gas law, and then integrate the equation from state 1 to state 2 (volume 1 to volume 2).

I remember that part now.
So I integrated and got ##\frac{nRT}{V}## I got ##nRT*ln(v)##
plugging in 10 for v I received 26.8621

if I plug that into the helmholtz eq, I get 268.8621. I'm not getting how to get to the next step.

 

[MexChemE]

I remember that part now.
So I integrated and got ##\frac{nRT}{V}## I got ##nRT*ln(v)##
plugging in 10 for v I received 26.8621

if I plug that into the helmholtz eq, I get 268.8621. I'm not getting how to get to the next step.

10 L is the net difference in volume, but when you integrate using limits, this is what you actually get
[tex]\int_{V_1}^{V_2} \frac{dV}{V} = \ln \frac{V_2}{V_1}[/tex]
You must integrate both sides of the fundamental equation, dA = -PdV.

 

[grandpa2390]

10 L is the net difference in volume, but when you integrate using limits, this is what you actually get
[tex]\int_{V_1}^{V_2} \frac{dV}{V} = \ln \frac{V_2}{V_1}[/tex]
You must integrate both sides of the fundamental equation, dA = -PdV.

...

I see. d'oh. that's just an I.D.10.T error. I plugged in the difference. I am supposed to solve it for each v and then subtract. Sometimes I feel like a nut. sometimes I don't

 

[grandpa2390]

10 L is the net difference in volume, but when you integrate using limits, this is what you actually get
[tex]\int_{V_1}^{V_2} \frac{dV}{V} = \ln \frac{V_2}{V_1}[/tex]
You must integrate both sides of the fundamental equation, dA = -PdV.

so I did that.
I'm getting 269. I'm still not doing something right.

turns out before. I somehow did the ln(v2/v1)... idk. but 268.621 is what I am getting when I plug the P into the helmholtz eq

if I plugged .083113 into R instead of .08205, I would get the right answer.
which is the same for the second part where I use the van der walls eq instead of an ideal gas.

this isn't the first two times either that that .083113 seems to have replaced .08205... is that supposed to be the universal gas constant... because it isn't exactly the same. and I thought the pressure had to be in pascals and the volume in cubic meters...
so it can't be that... unless it is a mistake

 

[MexChemE]

so I did that.
I'm getting 269. I'm still not doing something right.

turns out before. I somehow did the ln(v2/v1)... idk. but 268.621 is what I am getting when I plug the P into the helmholtz eq

You don't need to use any pressure in order to solve the problem, I think you may be using the equation the wrong way. This is what you should do
[tex]\int_{A_1}^{A_2} dA = - \int_{V_1}^{V_2} PdV = -nRT \int_{V_1}^{V_2} \frac{dV}{V}[/tex]
[tex]\Delta A = -nRT \ln \left( \frac{V_2}{V_1} \right)[/tex]
ΔA is what the problem is asking you to calculate.

 

[grandpa2390]

You don't need to use any pressure in order to solve the problem, I think you may be using the equation the wrong way. This is what you should do
[tex]\int_{A_1}^{A_2} dA = - \int_{V_1}^{V_2} PdV = -nRT \int_{V_1}^{V_2} \frac{dV}{V}[/tex]
[tex]\Delta A = -nRT \ln \left( \frac{V_2}{V_1} \right)[/tex]
ΔA is what the problem is asking you to calculate.

then what does the helmholtz energy eq matter for if you are not plugging anything into it, or using it?

 

[MexChemE]

then what does the helmholtz energy eq matter for if you are not plugging anything into it, or using it?

You are using it, you are integrating it. dA = - SdT - PdV is a differential statement of the equation, so in order to get something meaningful from it you must integrate it. Plugging in a "value" of dV to get a "value" of dA for the initial and final states of the process and then substract them makes no sense, that's what integration is for. When you integrate, you are accounting for the volume and pressure differences between the initial and final states, and that's all you need. The ΔA equation is still Helmholtz's equation, dA = -SdT - PdV is just a more general, differential form of it.

 

[grandpa2390]

You are using it, you are integrating it. dA = - SdT - PdV is a differential statement of the equation, so in order to get something meaningful from it you must integrate it. Plugging in a "value" of dV to get a "value" of dA for the initial and final states of the process and then substract them makes no sense, that's what integration is for. When you integrate, you are accounting for the volume and pressure differences between the initial and final states, and that's all you need. The ΔA equation is still Helmholtz's equation, dA = -SdT - PdV is just a more general, differential form of it.

so we don't really use it, we just see it and integrate whatever we are given for the gas (ideal gas, vanderwaal, etc) as a function of P in respect to V, and then I assume a function of S in respect to T (in cases where S and dT are not equal to zero)

 

[MexChemE]

so we don't really use it, we just see it and integrate whatever we are given for the gas (ideal gas, vanderwaal, etc) as a function of P in respect to V, and then I assume a function of S in respect to T (in cases where S and dT are not equal to zero)

Well, to me that counts as "using" it, but let's not argue semantics. You got the idea now, that's kind of why most thermodynamic definitions have a differential form, so we can integrate them and get useful results from them.

 

[grandpa2390]

Well, to me that counts as "using" it, but let's not argue semantics. You got the idea now, that's kind of why most thermodynamic definitions have a differential form, so we can integrate them and get useful results from them.

I'm sorry. I didn't mean that. i meant that it was more like a template. not plugging in value for each thing. but using it to find values.

 

[MexChemE]

I'm sorry. I didn't mean that. i meant that it was more like a template. not plugging in value for each thing. but using it to find values.

Yes, differential forms of thermodynamic equations are general statements, which you adapt to an specific process with the help of calculus and the conditions of the process itself.