# Volume Ratio of Carnot Engine

Consider a Carnot engine operating between a hot reservoir with temperature 700 °C and a cold reservoir with temperature 70 °C. The engine contains 0.13 mol of gas, and outputs a net 630 J of work.

Let state 1 be the state of the gas in the cycle when its volume is lowest. Determine the ratio $$V_2/V_1$$

I used the equation for work done by an isothermic process to determine the ratio of volumes

$$W=nRT\ln \frac{V_2}{V_1}$$ Since state one would be at its lowest volume, the temperature would have to be at its highest so that would be the temperature to be used in the equation in kelvins. $$630 = 0.13\times8.314\times(273+700)\ln\frac{V_2}{V_1}$$ $$\frac{630}{0.13\times8.314\times(273+700)}=\ln\frac{V_2}{V_1}$$ then exponentiating each side by $$e$$, we get $$e^\frac{630}{0.13\times8.314\times(273+700)}=\frac{V_2}{V_1}=1.82041$$ Is my process correct? As I'm getting that this answer is wrong.

• The 630 J is not the work done by the working gas when it is in contact with the hot reservoir. It is the net work done, which is the work done by the working gas when it is in contact with the hot reservoir minus the work done on the working gas when it is in contact with the cold reservoir. – Chet Miller Mar 22 at 2:43
• Thank you, I forgot that there would also be negative work being done on the second isothermic process – Sir Dhaliwal Mar 22 at 3:17