How do I calculate work and heat in a PV diagram for an ideal monoatomic gas?

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To calculate work and heat in a PV diagram for an ideal monoatomic gas, the ideal gas law (PV=nRT) is essential for determining temperatures. Work done by the gas can be calculated using specific formulas for isobaric and isothermal processes, with the total work being the sum of contributions from each segment of the path. The internal energy change for a non-cyclic process from A to C is not zero, requiring careful calculation of heat absorbed by the gas. The specific heat for an ideal monoatomic gas is (3/2)R, which aids in determining changes in internal energy. Understanding the specific processes involved is crucial for accurate calculations.
Muthumanimaran
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Homework Statement


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Homework Equations


Ideal gas law:
$$PV=nRT$$

Workdone by the gas:
For Isobaric process
$$W=P(V_{2}-V_{1})$$

For Isothermal Process
$$W=nRT \ln(V_{2}{/}V_{1})$$

First law:
$$U=Q-W$$

The Attempt at a Solution


subpart (a) is trivial, all we have to do is to substitute the given data in Ideal gas law to get the temperatures.

I have conceptual difficulty in part (b) and (c)only
The PV diagram conists of two Isothersms (DA and BC) and two Isobars (AB and CD)
How to calculate workdone by the gas?
Is it just the sum of the workdone by four processes ?

Next, Heat absorbed by gas?
Since the total Internal energy U of a cyclic process is zero, is it just equal to the total workdone?
 

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Muthumanimaran said:
How to calculate workdone by the gas?
Is it just the sum of the workdone by four processes ?
No. You are told that the system goes from A to C along path ABC. Which processes does that involve?
Muthumanimaran said:
Next, Heat absorbed by gas?
Since the total Internal energy U of a cyclic process is zero, is it just equal to the total workdone?
Is this a cyclic process?
(By the way, your statement is wrong in any case. The total internal energy change, ΔU, of a cyclic process is zero. It is vitally important to distinguish between thermodynamic quantities and changes in those quantities. Many mistakes are made here.)
 
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I think the focus of this problem is only the portion of the path between A and C. From the information given, you also know the volume at B. What is that volume?
 
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Oh, I got it now. I just need to calculate the work done from A to B and B to C and sum it up? Am I right?
 
Muthumanimaran said:
Oh, I got it now. I just need to calculate the work done from A to B and B to C and sum it up? Am I right?
That's my interpretation of the problem statement. Arguably, it is not very clear.
 
From A to C it is not a cyclic process, so the change in internal energy cannot be zero (from A to C), then how to find the heat absorbed by the gas?
 
What is the pressure, volume, and temperature of the gas at C? What is the pressure, volume, and temperature of the gas at B?
 
P=1 atm V=20lit at point C and T=PV/R

So T=20/R K at point C

Pressure at B is 4 atm T = 20/R
So V=5 lit
 
Without knowing the specific heat of the gas how do I calculate the change in internal energy from A to B? Workdone = P(VB-VA), is it just enough to plug these values to find the heat transfer from A to B? for B to C is an isotherm, so workdone is equal to heat right?
 
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Muthumanimaran said:
P=1 atm V=20lit at point C and T=PV/R

So T=20/R K at point C

Pressure at B is 4 atm T = 20/R
So V=5 lit
You do know the value of R, correct.? Please provide actual numbers for the temperatures at A, B, and C.
 
  • #11
Muthumanimaran said:
Without knowing the specific heat of the gas how do I calculate the change in internal energy from A to B? Workdone = P(VB-VA), is it just enough to plug these values to find the heat transfer from A to B? for B to C is an isotherm, so workdone is equal to heat right?
You are told that the gas is an ideal monoatomic gas. What does that tell you about its specific heat?
 
  • #12
Chestermiller said:
You are told that the gas is an ideal monoatomic gas. What does that tell you about its specific heat?
Specific heat is (3/2)R. Now I got it.
 

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