# A Three-Step Gas Cycle

by CurtisB
Tags: cycle, threestep
 P: 43 Hello, (1) For monatomic ideal gas, the internal energy can be also writen as $$U=\frac{3}{2}PV$$ The first step is an isochor process, no work is done, the heat is just the change of internal energy. You can use the formula and try again. (2) For the second process, it is an isothermal one, which means $$PV=nRT=\text{constant}$$ Therefore $$AP_1\times V_1=P_1\times V_f\Rightarrow V_f=AV_1$$ Substitute this relation into your formula and get it. (3) Actually, i don't completely understand why you think from your words. But i know the work is the area under the curve on the P-V diagram. The work done by the gas in the third step is $$-P_1(AV_1-V_1)$$. (The work done by the gas in the first step is $$0$$ since no area.) The total work is the area inside the colse process on the P-V diagram. It is $$0+AP_1V_1\ln{A}-P_1(AV_1-V_1)$$. Hope these are helpful. Goodluck.
 P: 43 A Three-Step Gas Cycle Hello, Part(1) You got it. Part(2) The formula $$W=pV\ln\left(\frac{V_f}{V_i}\right)$$ was derived in an isothermal process for ideal gas. In such an isothermal process, the product $$pV$$ is a constant, which means $$pV=p_iV_i=p_fV_f$$. The $$pV$$ substituted into the formula should be a product of some point on the process. $$p_1V_1$$ is not a point in the isothermal process, it's just a point in another process. Regards