How can I calculate the work done in a gas compression problem?

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Homework Help Overview

The discussion revolves around calculating the work done in a gas compression problem, specifically involving an ideal diatomic gas. The original poster attempts to determine the change in thermal energy and work done during the process, referencing relevant equations and attempting calculations based on pressure and temperature changes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of thermal energy and the need for the specific heat capacity (Cv) of a diatomic gas. Questions arise regarding the general equation for work done in a P-V context and the methods to calculate it, including integration and graphical interpretation.

Discussion Status

Some participants provide guidance on the value of Cv for diatomic gases and clarify the relationship between the integral of pressure and volume and the area under the P-V diagram. There is an acknowledgment of the original poster's confusion regarding the work calculation methods, with suggestions to explore further in upcoming course material.

Contextual Notes

Participants note that the original poster is uncertain about the specific heat capacity and the methods for calculating work, indicating a potential gap in the course material that will be addressed in future chapters.

BrainMan
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Homework Statement


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Homework Equations


ΔEth = nCvΔT

W = -∫ p dV

pV = nRT

The Attempt at a Solution


I tried to find the total change in thermal energy by finding the temperature change.

I first found the initial pressure

P1 = nRT/V = 32464.4 Pa

Then I solved for the final pressure

p1V12 = p2V22

p1V12 / V22 = p2 = 292179.6 Pa

Then I found the final temperature

T2 = pV/nR = 879 K

What I wanted to do next was use ΔEth = nCvΔT to find the total thermal energy and then subtract the work to find the heat. Unfortunately, I don't know what Cv is equal to or how to find the work.
 
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What is Cv for a diatomic gas? What is the general equation for the amount of P-V work done on the surroundings?
 
Chestermiller said:
What is Cv for a diatomic gas? What is the general equation for the amount of P-V work done on the surroundings?
Cv is one the things that is confusing me. In my book it has a table for a couple of Cv's for different gases so I'm not sure which to use.

https://lh3.googleusercontent.com/Rv3s5lQ2vyLKOnlbAg64OHbFcNAw_pmlo3t_v1YoFHulQit7hmXE3QNfhnkkY2CkbFphumXv2Fur1BgU6j7CJ2ndcsMKnvvKsKTcS0La7W1cMEUdeQQ3FbwVGqP5uWtq6m-kac0EcRYm_z-pHDVDPFX_KTcGziQOZQlpRX-c_lkY1xWJEPMQcEwFIdgZ7kYoCr9xMVr0IzzS4h_1xe3yKY4GOsox4ARm5AfFURq_crZHRoHdsGH9XYU8hyrFe2AFoDavbMAn2ee_h3yrviHDKjvCau1GbRp6K7xKT5ql1j64izFwZO32BxrpXCMhI5CEMjMnIc5hN6-259RsF8SsgIiZxsnQD7TAu3-9d8Xj3vMR35dMFtUUSelQQqWnQblf0ylCU0DYrzG9N2Mfx3g3_WlsPUZkkoEKP9dmSO8HU0KEfev7mT2Xz8cB2mcGLaFXeywOdi0kWonRkAfn1XG3mGgT2S4j0GWWMPL-AZpMb5RgzYHMe_HFK0oKB18wEPGsX3PAGwTh8yoklqY_TLssNCRJeLLKTYIhXf1Pl7LHumgHEHRrhfuHt0ZtmZ5AEQKq64Nooh5rsHMRfAGXyvQP6SdsadwWXxxc2evTNwlvnTGZRg9ucT4P=w619-h662-no
 
For an ideal diatomic gas, the molar Cv is equal to 5R/2. Did they not cover this in your course?
 
Chestermiller said:
For an ideal diatomic gas, the molar Cv is equal to 5R/2. Did they not cover this in your course?

I looked ahead and it seems it will be covered in the next chapter. OK so that solves that problem. What I'm wondering about now is the work. My book wasn't very clear about work and the two ways it gave to calculate work were to W = -∫ p dV or to find the area under the curve of the pV diagram.
 
BrainMan said:
I looked ahead and it seems it will be covered in the next chapter. OK so that solves that problem. What I'm wondering about now is the work. My book wasn't very clear about work and the two ways it gave to calculate work were to W = -∫ p dV or to find the area under the curve of the pV diagram.
The equation you wrote for the work represents the work done by the surroundings on the system. This equation is correct. You are aware that the integral of PdV is the same as the area under the curve of the pV diagram, correct? In this problem, to get the work, you are going to have to integrate pdV.
 

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