The reversible expansion of an ideal gas unde | vedclass

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Question

(C) For an isothermal process,the temperature remains constant. Therefore,$T_1 = T_2$ is correct.$(B)$ In an adiabatic expansion,the gas does work at

Vedclass · Accepted Answer

$(AC)$

Vedclass · Answer

(C) For an isothermal process,the temperature remains constant. Therefore,$T_1 = T_2$ is correct.$(B)$ In an adiabatic expansion,the gas does work at the expense of its internal energy,leading to a decrease in temperature. Thus,$T_3  T_1$ is incorrect.$(C)$ The area under the $P-V$ curve represents the work done. For the same final volume $V_2$,the area under the isothermal curve is greater than the area under the adiabatic curve. Thus,$W_{	ext{isothermal}} > W_{	ext{adiabatic}}$ is correct.$(D)$ For an isothermal process,$\Delta U = 0$. For an adiabatic expansion,$\Delta U  -	ext{ve}$,$\Delta U_{	ext{isothermal}} > \Delta U_{	ext{adiabatic}}$ is correct.Therefore,statements $(A)$,$(C)$,and $(D)$ are correct.

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