Understanding the PV Diagram of a Monatomic Gas

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Homework Help Overview

The discussion revolves around analyzing the PV diagram of a monatomic gas, focusing on calculating changes in internal energy, heat, and work for various processes and the overall cycle. The context involves thermodynamic principles and equations related to ideal gases.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate work and heat for different processes using the PV diagram and relevant equations. Some participants question whether the work calculated is the work done by the gas and suggest verifying unit consistency.

Discussion Status

Participants are actively discussing the calculations presented, with some confirming the original poster's approach while others point out potential issues with unit conversions. There is no explicit consensus on the correctness of the calculations, but guidance on checking units has been provided.

Contextual Notes

There is a noted discrepancy regarding the units used for pressure and volume, as the original poster used atmospheres and litres instead of the MKS system. This raises questions about the accuracy of the energy, heat, and work calculations derived from those units.

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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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