Question

An ideal gas is allowed to expand both adiabatic reversibly and adiabatic irreversibly in an isolated system. If $T_i$is the initial temperature and $T_f$ is the final temperature, which of the following statement is correct?

• A. $T_f>T_i$ for reversible process but $T_f=T_i$ for irreversible process
• B. ${\left(T_f\right)}_{rev}={\left(T_f\right)}_{irrev}$
• C. $T_f=T_i$ for both reversible and irreversible processes}
• D. ${\left(T_f\right)}_{irrev}>{\left(T_f\right)}_{rev}$

Answer 1

The correct option is A $T_f>T_i$ for reversible process but $T_f=T_i$ for irreversible process

According to first law of thermodynamics

$\Delta Q=\Delta U+\Delta W$

An isolated system is adiabatic. This means $\Delta Q=0$. The first law in this case yields

$0=\Delta U+\Delta W=>\Delta W=-\Delta U$ … … (1)

For expansion, $\Delta W$ is positive and hence $\Delta U$ is negative. This means $T_f$ is less than $T_i$ in both the cases. However it is interesting to see in which case the final temperature is greater.

For the same expansion of volume, the work done in irreversible process is greater than that in reversible one because the system has to work against friction etc. Thus

$\Delta W_{irreversible}>\Delta W_{r}eversible$

=>$-\Delta U_{irreversible}>-\Delta U_{reversible}$ (from equation (1))

=>$\Delta U_{irreversible}<\Delta U_{reversible}$

=>$\Delta T_{irreversible}<\Delta T_{reversible}$

=>$T_{f-irreversible}>T_{f-reversible}$