Monatomic Gas and Isotherm - W, U Calculations

[dvolpe]

Homework Statement

. Suppose a monatomic ideal gas is changed from state A to state D by one of the processes shown on the PV diagram below, in which each interval on the vertical (P) axis corresponds to 2 atm, and the values of the lines on the horizontal (V) axis equal 1 L, 2 L, and 4 L.

Graph attached.


(a) Find the total work done on the gas if it follows the constant volume path A-B followed by the constant pressure path B-C-D.
(b) Find the total change in internal energy and total heat flow into the gas.

Homework Equations

For isotherm W = nRt (ln (Vf/Vi)
PV = nRT

The Attempt at a Solution

From A to B to C is an isotherm. Use equation above but substitute PV for nRT for either A or C as they are the same - got 255.6 J. Then add in PdeltaV for C to D as on 2 different isotherms. Total work = A to C isotherm work plus C to D work = 655.6. HELP!

 

[RTW69]

The path is A-B-C-D. Not sure why you are interested in isotherms. The work from A-B is 0 since there is no change in volume. The work from B-C-D is just the area= P*change in volume. Do you have an correct answer for this problem?

 

[dvolpe]

No. But in a similar problem with intervals on y-axis to be 1 atm and values of lines on x-axis to be 5, 10, and 20 L, the answer is -1520 Joules. I can't get this answer by adding the areas under B-C and C-D to get work (A-B is 0 since it is isochoric). Any ideas?

 

[RTW69]

Yes you can. Look at the example problem. The area under B-C-D is the work.

Work= P*delta V= (1 atm) *(15 l) (101325 Pa/atm) * (.001 M^3/l)= 1520 J. I am confused why there is a negative work since the change in volume is positive so the work done is positive.

 

[Andrew Mason]

Yes you can. Look at the example problem. The area under B-C-D is the work.

Work= P*delta V= (1 atm) *(15 l) (101325 Pa/atm) * (.001 M^3/l)= 1520 J. I am confused why there is a negative work since the change in volume is positive so the work done is positive.

The work done by the gas would be 1520 J. in that example. But the question asks for the work done ON the gas.

AM

 

[dvolpe]

Thanks..I got the 1520 in that other example by using the 1.013 e 5number to convert (I had used 1.0 e 5 earlier). Now how do I find the internal energy and total heat flow of the total process? I know that change in internal energy = Q-W. If I use change in U = 3/2*nRdeltaT and substitute PV for nRdelta T at each step, I get 3039 J for change in internal energy but that is the answer for total heat flow into the gas. I am confused!

 

[Andrew Mason]

Thanks..I got the 1520 in that other example by using the 1.013 e 5number to convert (I had used 1.0 e 5 earlier). Now how do I find the internal energy and total heat flow of the total process? I know that change in internal energy = Q-W. If I use change in U = 3/2*nRdeltaT and substitute PV for nRdelta T at each step, I get 3039 J for change in internal energy but that is the answer for total heat flow into the gas. I am confused!

Careful:

\Delta U = nC_v\Delta T = \Delta (PV)C_v/R

You appear to be using PV not \Delta (PV).

AM