# Isothermic Work done by a gas

1. Jan 20, 2006

### Valhalla

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 $$p_1 = 8 atm$$ $$V_1 = .1m^3$$ $$p_2 = 1 atm$$$$V_2 = .8m^3$$?
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??!?!?

Last edited: Jan 20, 2006
2. Jan 20, 2006

### Staff: Mentor

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?

3. Jan 20, 2006

### Valhalla

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.