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

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SUMMARY

The discussion focuses on calculating work done during gas compression using the equations ΔEth = nCvΔT and W = -∫ p dV. The initial pressure (P1) was calculated as 32464.4 Pa, and the final pressure (P2) was determined to be 292179.6 Pa. The final temperature (T2) was found to be 879 K. For an ideal diatomic gas, the molar heat capacity at constant volume (Cv) is established as 5R/2, which is crucial for calculating thermal energy changes.

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  • Understanding of the ideal gas law (pV = nRT)
  • Knowledge of thermodynamic equations for work and energy (W = -∫ p dV)
  • Familiarity with heat capacities, specifically Cv for diatomic gases
  • Basic calculus for integration in thermodynamic contexts
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  • Study the integration process for calculating work done in thermodynamic systems
  • Explore the concept of heat capacities for different types of gases
  • Investigate graphical methods for analyzing P-V diagrams and calculating work
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Homework Statement


upload_2016-12-30_10-56-48.png


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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