Question

An ideal gas is allowed to expand both reversibly and 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. $(T_f{)}_{irrev.}>(T_f{)}_{rev.}$
• B. $T_f>T_i$ for reversible process by $T_f=T_i$ for irreversible process
• C. $(T_f{)}_{rev.}=(T_f{)}_{irrev.}$
• D. $T_f=T_i$ for both reversible and irreversible process

Answer 1

The correct option is A $(T_f{)}_{irrev.}>(T_f{)}_{rev.}$

According to law of thermodynamics, $\Delta E=\Delta q+\Delta w$

Since, the system is isolated, therefore the change takes place adiabatically. Thus, $\Delta q=0$

So, $\Delta E=\Delta w$

We know, $\Delta w_{irreversible}>\Delta w_{reversible}$

This implies, $\Delta E_{irreversible}>\Delta E_{reversible}$

$\Delta E=nR\Delta T$

As a result, $\Delta T_{irrev.}>\Delta T_{rev.}$

Thus, $(T_f{)}_{irrev.}>(T_f{)}_{rev.}$