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The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement is incorrect?

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$T_{3} > T_{1}$

$T_{1} = T_{2}$

$Δ U_{i s o t h e r m a l} > Δ U_{a d i a b a t i c}$

$W_{i s o t h e r m a l} > W_{a d i a b a t i c}$

answer is A.

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Detailed Solution

In an isothermal process, temperature is constant $T_{2} = T_{1}$

Area under P-V graph is more in isothermal $W_{i s o t h e r m a l} > W_{a d i a b a t i c} \Rightarrow N o w Δ U_{i s o t h e r m a l} = 0$

In adiabatic $T_{1} {V_{1}}^{γ - 1} = T_{3} {V_{3}}^{γ - 1} \Rightarrow \frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{3}})}^{γ - 1} . T_{3} < T_{1}$ hence internal energy decreases $Δ U_{i s o t h e r m a l} > Δ U_{a d i a b a t i c}$

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