Understanding the PV Diagram of a Monatomic Gas

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SUMMARY

The discussion focuses on calculating the change in internal energy, heat, and work for a monatomic gas using a PV diagram. Key equations include PV = nRT, ΔU = Q - W, and specific heat calculations for constant pressure and volume processes. The participant initially miscalculated work and energy due to incorrect unit usage, specifically mixing atmospheres and liters instead of converting to the MKS system. The correct work done during the process from B to C is 350 J, not 3.5 J.

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Homework Statement


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

hMivkKO.png


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):

3535rkr.png


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:

x45zO8h.png


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:

FW9DwiX.png


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:

ZiCNBTq.png


But I got this problem incorrect. What errors have I made?
 
Last edited:
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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
 

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