Finding Q, W, and ΔU for three processes in a monatomic gas cycle

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Homework Statement


Consider the following PV diagram of one mole of a monatomic gas:

PV diagram of a triangular cycle with vertices A at (1.0, 0.5), B at (1.0, 3.0) and C at (3.0, 0.5), with arrows showing the path A to B to C and back to A.


The units are in atm and litres.

Find the change in internal energy, heat, and work for all 3 processes and for the cycle as well.

Homework Equations



PV = nRT
ΔU = Q - W
ΔU = (3/2)*R*ΔT
Q = (5/2)*R*ΔT (for constant pressure)

The Attempt at a Solution



For the work, I just found the areas:

A -> B: 0 J
B -> C: +3.5 J
C -> A: -1 J

For the cycle, the work done is just the total work added:
3.5 - 1 = 2.5 J

For the cycle, the Q is the same as the Work:
2.5 J

For the cycle, the ΔU = 0.

So far I have this (unless I did something wrong):

Table with columns ΔU, Q and W and rows A -> B, B -> C, C -> A and Cycle; W values 0, 3.5 and -1 are filled in, the cycle row shows ΔU 0, Q 2.5 and W 2.5, and the other cells are blank.


Since for a constant volume process the ΔU = Q, I decided to do that A -> B first.

T_A = (0.5 atm)(1 L) = (1)(0.0821)T, T = 6.09 K
T_B = (0.5 atm)(1 L) = (1)(0.0821)T, T = 36.5 K

ΔT = 30.41 K

ΔU = (3/2)*R*ΔT = (3/2)*0.0821*30.41 = 3.75 J = Q

So:

Table of thermodynamic values with columns ΔU, Q and W and rows A -> B, B -> C, C -> A and Cycle. A -> B has ΔU 3.75, Q 3.75, W 0; B -> C has W 3.5; C -> A has W -1; Cycle has ΔU 0, Q 2.5, W 2.5.


The constant pressure process looks easier so I'll do C -> A next:

T_C = (0.5 atm)(3 L) = (1)(0.0821)T, T = 18.27 K
T_A = (0.5 atm)(1 L) = (1)(0.0821)T, T = 6.09 K

ΔT = -12.18

Q = (5/2)*R*ΔT = (5/2)*0.0821*(-12.18) = -2.5 J

ΔU = Q - W = (-2.5) - (-1) = -1.5

So:

Table of ΔU, Q and W values for processes A to B, B to C, C to A and the cycle; A to B has ΔU 3.75, Q 3.75, W 0; B to C has W 3.5; C to A has ΔU -1.5, Q -2.5, W -1; cycle has ΔU 0, Q 2.5, W 2.5.


Now I can just add up the unknown columns:

ΔU for B -> C: 3.75 + (-1.5) + x = 0, x = -2.25

Q for B -> C: 3.75 + (-2.5) + x = 2.5, x = 1.25

So I got:

Table of ΔU, Q and W values in joules for processes A to B, B to C, C to A, and the cycle; A to B has ΔU 3.75, Q 3.75, W 0.


But I got this problem incorrect. What errors have I made?
 
Last edited:
Your work looks correct. Are you sure the work asked is the work done by the gas? ehild
 
PhizKid said:
But I got this problem incorrect. What errors have I made?
Your method is correct but you have to check your units. In the MKS system the units for pressure are Pascals and the units for volume are m3. Your PV diagram is in atmospheres and litres.

So, for example, the work done by the gas from B to C is not 3.5 Joules but 3.5 atm. litres = 3.5(101325Pa) 10-3 m3 =350 J.

AM
 
AM is right. You used litres and atm as units, then you got also the energy, heat, and and work in units (Latm) instead of joules.

ehild