Discussion Forum

The reversible expansion on an ideal gas under adiabatic and isothermal conditions is shown in the figure.
Which of the following statement(s) is (are) correct?

Answer:

Explanation:

(a) Since, change of state ($p_{1},V_{1},T_{1}$) to ($p_{2}, V_{2},T_{2}$) is isothermal therefore $T_{1}=T_{2}$

(b) Since, change of state ($p_{1},V_{1},T_{1}$) to ($p_{3}, V_{3},T_{3}$)  is an adiabatic  expansion it brings about cooling of gas , therefore $T_{3} <T_{1}$

(c)   Work-done is the area under the  curve of p-V diagram' As obvious  from the given diagram, magnitude

of area under the isothermal curve  adiabatic curve , hence

$w_{isothermal}$   > $W_{adiabatic}$

Note: Here only magnitudes of work is being considered otherwise both works have negative sign.

(d)$\triangle U=nC_{v}\triangle T$

In isothermal Process, $\triangle U=0,$ as $\triangle T=0$ in adiabatic process

 $\triangle U=nC_{v}$ ($T_{3}-T_{1})<0$ as $T_{3}<T_{1}$

$\Rightarrow$   $\triangle U_{isothermal}$ >  $\triangle U_{adiabatic}$