Isothermic Work by a Gas: Linear Relationship or Integration?

AI Thread Summary
The discussion revolves around calculating the work done by a gas during isothermal expansion from given initial and final pressures and volumes. The original poster used integration to solve the problem, while their teacher suggested a linear relationship approach, which led to confusion. The poster's integration method, involving the formula W = nRT ln(V2/V1), was confirmed as correct by their professor, who acknowledged a mistake in the solutions guide. The key takeaway is that for isothermal processes, integration is necessary to accurately determine work done, as a linear relationship between pressure and volume does not apply. This highlights the importance of using the correct mathematical approach in physics problems involving gases.
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Work done by a gas; Isothermic situation

A gas expands isothermically from p_1 , V_1 to p_2 , V_2 . What is the work done by the gas if p_1 = 8 atm V_1 = .1m^3 p_2 = 1 atmV_2 = .8m^3?
I solved the problem using the integral P*dv but my teacher posted the solutions this evening and drew a linear relationship between the two and solved that way. The problem as you see above is how it was given to us verbatim. Should I have assumed that there was a linear relationship between the two? This is Physics with Calc so it isn't like we can't do the integration. Here was my solution
W=pdv=\int_ {V_1}^{V_2}\frac{nRTdV}{V}
nRT=p_1V_1
nRT\ln\frac{V_2}{V_1}
p_1V_1\ln\frac{V_2}{V_1}
Then I just plugged in my values and got an answer. I guess my real question is why should I have assumed there was a linear relationship here?
 
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Your solution looks correct to me. I don't see how the correct answer could be obtained by assuming a linear relationship (between P and V, I presume) where one doesn't exist. Did he get the same answer as you?
 
Thanks for your input Doc! Just got a reply from my professor. He said he made a mistake when writing the solutions guide and verified my answer.
 
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