Heat absorbed by an ideal gas in a cycle

Thread starter Isomorphism
Start date Nov 13, 2016
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Cycle First law of thermodynamics Gas Heat Heat absorbed Ideal gas

AI Thread Summary

In the discussion about heat absorption in an ideal gas cycle, it is established that process AC is adiabatic, resulting in zero heat absorption. During process CB, the heat absorbed is calculated using the formula involving specific heat at constant pressure and a change in temperature. For process BA, the heat absorbed is determined using the ideal gas law and logarithmic relationships. Despite these calculations, the total heat absorbed does not match any provided options, leading to confusion about the accuracy of the results. The consensus is that the calculations appear correct, suggesting a potential issue with the options given.

Nov 13, 2016

Isomorphism

Homework Statement

Homework Equations

and the attempt at a solution:[/B]
AC is adiabatic and AB is isothermal.

Heat absorbed during process AC = 0 (adiabatic).
Heat absorbed during process CB = C_p \triangle T=-\frac{\gamma}{\gamma -1} (P_2V_3 - P_2V_2)
Heat absorbed during process BA = RT\ln \frac{V_3}{V_1}

Summing it up, does not give any of the options. What went wrong?

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Nov 13, 2016

Chestermiller

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I agree with your answer. In my judgment nothing went wrong.

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