Question

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

• A. $T_1=T_2$
• B. $T_3>T_1$
• C. $w_{isothermal}>w_{adiabatic}$
• D. ${\Delta U}_{isothermal}>{\Delta U}_{adiabatic}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $T_1=T_2$
C $w_{isothermal}>w_{adiabatic}$
D ${\Delta U}_{isothermal}>{\Delta U}_{adiabatic}$

Since the process is an isothermal . so $\Delta T=0{⟹T}_1=T_2$

The area under the P-V graph denotes the work done by the system. The area is more for an isothermal process. so $w_{isothermal}{>w}_{adiabatic}$

In an isothermal process, change in internal energy is zero. So $\Delta U_{isothermal}=0$

In adiabatic process, heat energy released will be zero $Q=0$

From first law,

$Q=w+\Delta U$

${\Delta U}_{adiabatic}{=-w}_{adiabatic}<0$.

So ${\Delta U}_{isothermal}{>\Delta U}_{adiabatic}$

The curve which downward has the less temperature in $P-V$ graph of adiabatic expansion. so $T_1>T_3$