Isothermic and adiabatic compression

[mk9898]

Homework Statement

Determine the ratio of the work that you must expend when a single-atom ideal Gas once isothermic and once adiabatic to 1/10 of its original volume is compressed. In both cases the initial pressure, initial volume and gas amount are all equal.

The Attempt at a Solution

Case 1: Isotherm --> $\triangle Q = W$
(saving derivation) $Q = RTln(10)$

Case 2: Adiabatic: --> $\frac{fNk\triangle T}{2T}= \frac{-NkdV}{V}$

$\frac{f}{2}lnT + ln(10) = constant$

 

[Andrew Mason]

Homework Statement

Determine the ratio of the work that you must expend when a single-atom ideal Gas once isothermic and once adiabatic to 1/10 of its original volume is compressed. In both cases the initial pressure, initial volume and gas amount are all equal.

The Attempt at a Solution

Case 1: Isotherm --> $\triangle Q = W$
(saving derivation) $Q = RTln(10)$

Ok.

Case 2: Adiabatic: --> $\frac{fNk\triangle T}{2T}= \frac{-NkdV}{V}$

$\frac{f}{2}lnT + ln(10) = constant$

You will have to explain your thinking there. What is "f"? What is the formula you are using for adiabatic work? Start with:
$$W = \int_{V_0}^{V_f} PdV$$

 

[Dr Dr news]

Case 1. Isothermal compression, pV = con, Wk = ∫ p dV, Case 2. Adiabatic compression, p(V)^γ = con, γ = c(p)/c(V), Wk = ∫ p dV

 

[mk9898]

Thanks for the replies.

Case 2: $W=-\triangle U$
$\triangle U =nC_v(T_2-T_1)$ $(C_v$ is with the mole of the gas)
Therefore $W = nC_v(T_1-T_2)$>0

Now I'm scratching my head at the next step... If I solve the integral for W then I get what I got from case 1 which I believe doesn't help me... any thoughts?

 

[Dr Dr news]

In case 1. Wk = ∫ p dV = ∫ con1. (dV / V); in case 2. Wk = ∫ p dV = ∫ con2. (dV / V^γ)

 

[Andrew Mason]

Thanks for the replies.

Case 2: $W=-\triangle U$
$\triangle U =nC_v(T_2-T_1)$ $(C_v$ is with the mole of the gas)
Therefore $W = nC_v(T_1-T_2)$>0

Now I'm scratching my head at the next step... If I solve the integral for W then I get what I got from case 1 which I believe doesn't help me... any thoughts?

For the adiabatic part, you can use the formula for adiabatic work (see, for example, this hyperphysics page )

or you can do this: Using the first law, Q = ΔU + W = C_vΔT + W, so W = Q - C_vΔT. Use the adiabatic condition to find the change in temperature (hint: in PV^γ=K, substitute RT/V for P to get: TV^(γ-1) = constant).

AM

 

[mk9898]

Ok so I have $W = \frac{K((\frac{V}{10})^{1-\gamma}-(V)^{1-\gamma})}{1-\gamma}$. The exercise wants us to determine the ratio i.e. compare the two values. Maybe I'm not seeing something but such a comparison doesn't lead to any easy cancellation of determines between $RTln(10)$ and $\frac{K((\frac{V}{10})^{1-\gamma}-(V)^{1-\gamma})}{1-\gamma}$.

 

[Dr Dr news]

What is K? Once you insert K into the W equation you should see how to simplify the ratio.

 

[mk9898]

Ah ok then I have: $\frac{\frac{RT9/10}{1-\gamma}}{RTln(10)}$ which is $\frac{0.9}{ln10(1-\gamma)}$ and then I can solve for gamma (gamma is 1+f/2 = 5/2). Is my logic right there?

 

[Andrew Mason]

Ok so I have $W = \frac{K((\frac{V}{10})^{1-\gamma}-(V)^{1-\gamma})}{1-\gamma}$. The exercise wants us to determine the ratio i.e. compare the two values. Maybe I'm not seeing something but such a comparison doesn't lead to any easy cancellation of determines between $RTln(10)$ and $\frac{K((\frac{V}{10})^{1-\gamma}-(V)^{1-\gamma})}{1-\gamma}$.

For the adiabatic case: $T_f/T_i = (V_i/V_f)^{γ-1}$. From that find ΔT and just plug that into the first law W = Q-ΔU to find W. (hint: ΔU = C_vΔT . What is Q? C_v?)
Compare that to RT_iln(10).

AM

 

[mk9898]

Thanks for the post. Q = 0 since it's an adiabatic process i.e. no change in heat in the system. $\triangle T = T_f - T_i >0$ since it is an adiabatic compression. Or did you mean something else with finding $\triangle T$?

 

[Andrew Mason]

Thanks for the post. Q = 0 since it's an adiabatic process i.e. no change in heat in the system. $\triangle T = T_f - T_i >0$ since it is an adiabatic compression. Or did you mean something else with finding $\triangle T$?

So work it out: $\Delta T = T_i(\left(\frac{V_i}{V_f}\right)^{\gamma-1} -1) = ??$

AM

 

[mk9898]

And then is $W = C_vT_i(1-\left(\frac{V_i}{V_f}\right)^{\gamma-1})$

 

[Andrew Mason]

And then is $W = C_vT_i(1-\left(\frac{V_i}{V_f}\right)^{\gamma-1})$

Correct. In the statement of the first law that I used, Q = ΔU + W, W is the work done BY the gas. So W is negative.

All you have to do is express C_v in terms of R and then divide by the isothermal work.

AM

 

[mk9898]

$W = C_vT_i(1-\left(\frac{V_i}{V_f}\right)^{\gamma-1}) = \frac{f}{2}RT_i(1-\left(\frac{V_i}{V_f}\right)^{\gamma-1}) = \frac{f}{2}RT_i(1-10^{\gamma-1})$
Isotherm should not be "T" above but $RT_iln(10)$
Then I take a look at the ratio.
Thanks a lot for the patience Andrew and for the help!

 

[mk9898]

Correct. In the statement of the first law that I used, Q = ΔU + W, W is the work done BY the gas. So W is negative.

Shouldn't that be, that work is being done TO the gas? If the work was being done by the gas, then it would be positive since the gas would be doing the work and hence pushing the volume outward? Or is this just based on perspective?

 

[Andrew Mason]

Shouldn't that be, that work is being done TO the gas? If the work was being done by the gas, then it would be positive since the gas would be doing the work and hence pushing the volume outward? Or is this just based on perspective?

W is the work done BY the gas. Another way to write the first law is: ΔU = Q-W. This says that the change in internal energy of the system is equal to the heat flow (into) the system minus the work done BY the system. In an isothermal process with an ideal gas, ΔU = 0. So the work done BY the system is equal to the heat flow into the system.

AM

 

[Dr Dr news]

Another thought is for an isothermal process, as you take work out you must add an equal amount of heat energy to keep T constant and conversely as you put work in say in compressing the gas you need to transfer heat energy out to keep T constant.