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##

Homework Helper

## 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:
$$W = \int_{V_0}^{V_f} PdV$$

mk9898
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)

Homework Helper
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 = CvΔT + W, so W = Q - CvΔT. Use the adiabatic condition to find the change in temperature (hint: in PVγ=K, substitute RT/V for P to get: TV(γ-1) = constant).

AM

mk9898
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
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?

Homework Helper
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 = CvΔT . What is Q? Cv?)
Compare that to RTiln(10).

AM

mk9898
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##?

Homework Helper
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
mk9898
And then is ##W = C_vT_i(1-\left(\frac{V_i}{V_f}\right)^{\gamma-1}) ##

Homework Helper
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 Cv in terms of R and then divide by the isothermal work.

AM

mk9898
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!

Last edited:
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?