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

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To calculate work done in a gas compression problem, the relevant equations include ΔEth = nCvΔT and W = -∫ p dV. The initial and final pressures were calculated, leading to a final temperature of 879 K. For a diatomic gas, the molar Cv is 5R/2, which will be covered in the next chapter. The work can be determined by integrating pressure with respect to volume or by finding the area under the pV diagram. Understanding these concepts is crucial for solving the problem effectively.
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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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