Finding Heat and Work in a Three-Step Gas Cycle

[CurtisB]

Homework Statement

A monatomic ideal gas has pressure p_1 and temperature T_1. It is contained in a cylinder of volume V_1 with a movable piston, so that it can do work on the outside world.

Consider the following three-step transformation of the gas:

1. The gas is heated at constant volume until the pressure reaches Ap_1 (where A >1).
2. The gas is then expanded at constant temperature until the pressure returns to p_1.
3. The gas is then cooled at constant pressure until the volume has returned to V_1.

It may be helpful to sketch this process on the pV plane.

Part 1-
How much heat DeltaQ_1 is added to the gas during step 1 of the process?
Express the heat added in terms of p_1, V_1, and A.

Part 2-
How much work W_2 is done by the gas during step 2?
Express the work done in terms of p_1, V_1, and A.

Part 3-
How much work W_3 is done by the gas during step 3?
If you've drawn a graph of the process, you won't need to calculate an integral to answer this question.
Express the work done in terms of p_1, V_1, and A.

Homework Equations

C_V = 12.47
R = 8.31

The Attempt at a Solution

Part 1-
I tried Q = p_1*V_1*(C_V/R) = 1.5*Ap_1*V_1 but I was told this is the final internal energy, not the change in internal energy. so I worked out that

Q = [1.5*p_1*V_1*(AT_1-T_1)] / T_1 but the answer does not depend on AT_1 or T_1

Part 2-
all I've got so far is
W = nRT*ln(V_f/V_i) = pV*ln(V_f/V_i)
but that's about as far as I get.

Part 3-
I got Ap_1*V_1 but this is what the value would be if it were coming from V = 0. So I re-arranged pV=nRT to eventually get

W = p_1[(p_1V_1)/(Ap_1) - V_1]
but this is also wrong how do I take into account the initial state, wouldn't I just be able to write W = (Ap_1V_1) - V_1 ?
Could someone please set me on the right path, I have been up late each night this week trying to work this out.

 

[variation]

Hello,
(1)
For monatomic ideal gas, the internal energy can be also written as U=\frac{3}{2}PV
The first step is an isochor process, no work is done, the heat is just the change of internal energy.
You can use the formula and try again.
(2)
For the second process, it is an isothermal one, which means PV=nRT=\text{constant}
Therefore AP_1\times V_1=P_1\times V_f\Rightarrow V_f=AV_1
Substitute this relation into your formula and get it.
(3)
Actually, i don't completely understand why you think from your words.
But i know the work is the area under the curve on the P-V diagram.
The work done by the gas in the third step is -P_1(AV_1-V_1).
(The work done by the gas in the first step is 0 since no area.)
The total work is the area inside the colse process on the P-V diagram.
It is 0+AP_1V_1\ln{A}-P_1(AV_1-V_1).

Hope these are helpful. Goodluck.

 

[CurtisB]

So for part one, if delta_U = 3/2(pV)
then delta_U should equal 3/2* the change in p * V = 3/2(Ap_1 - p_1)V_1. does this look right.

for part 2, W = pV*ln(V_f/V_i)
then W = p_1V_1*ln(AV_1/V_1), or should p and V be the changes in p and V, the feedback I get when I tried that answer was that N*k_B*T = A*p_1*V_1

 

[variation]

Hello,

Part(1)
You got it.

Part(2)
The formula W=pV\ln\left(\frac{V_f}{V_i}\right) was derived in an isothermal process for ideal gas.
In such an isothermal process, the product pV is a constant, which means pV=p_iV_i=p_fV_f.
The pV substituted into the formula should be a product of some point on the process.
p_1V_1 is not a point in the isothermal process, it's just a point in another process.

Regards

 

[xmonsterx]

im working on the same problem now but I am still a little confused. so what is the answer to part a,b,c.

part a = 3/2(Ap1-p1)V1 ?

part b = ??

part c = ??


im so lost! can anyone please help me? thanks a bunch!