Kruthik
Last Activity: 6 Years ago
I think 1/loge2 is the answer. Let’s see how.
Let Pi,Ti, and Vo be the initial pressure, temperature, and volume.
Consider the isobaric expansion.
Work done in isobaric expansion is PΔV, so W=Pi(2Vo – Vo). = Pi(Vo).
Now let’s consider the isothermal expansion.
Work done in isothermal process is given by, W = nRTloge(Vi/Vf).
So, W’ = nRTiloge(2Vo/Vo) = nRTiloge2.
By ideal gas law, nRTi = PiVo.
Thus, W’=PiVologe2.
Hence, W/W’ = 1/loge2.
